steps in problem solving involving factors of polynomials

4.4 Solve Polynomial Equations by Factoring

Learning objectives.

  • Review general strategies for factoring.
  • Solve polynomial equations by factoring.
  • Find roots of a polynomial function.
  • Find polynomial equations given the solutions.

Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials:

  • Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Determine the number of terms in the polynomial.

  • Factor four-term polynomials by grouping.
  • Factor trinomials (3 terms) using “trial and error” or the AC method.

Factor binomials (2 terms) using the following special products:

  • Look for factors that can be factored further.
  • Check by multiplying.

Note : If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization. In addition, not all polynomials with integer coefficients factor. When this is the case, we say that the polynomial is prime.

If an expression has a GCF, then factor this out first. Doing so is often overlooked and typically results in factors that are easier to work with. Furthermore, look for the resulting factors to factor further; many factoring problems require more than one step. A polynomial is completely factored when none of the factors can be factored further.

Factor: 54 x 4 − 36 x 3 − 24 x 2 + 16 x .

This four-term polynomial has a GCF of 2 x . Factor this out first.

54 x 4 − 36 x 3 − 24 x 2 + 16 x = 2 x ( 27 x 3 − 18 x 2 − 12 x + 8 )

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

steps in problem solving involving factors of polynomials

Answer: 2 x ( 3 x − 2 ) 2 ( 3 x + 2 ) . The check is left to the reader.

Factor: x 4 − 3 x 2 − 4 .

This trinomial does not have a GCF.

x 4 − 3 x 2 − 4 = ( x 2             ) ( x 2             ) = ( x 2 + 1 ) ( x 2 − 4 )          D i f f e r e n c e   o f   s q u a r e s = ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

The factor ( x 2 + 1 ) is prime and the trinomial is completely factored.

Answer: ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

Factor: x 6 + 6 x 3 − 16 .

Begin by factoring x 6 = x 3 ⋅ x 3 and look for the factors of 16 that add to 6.

The factor ( x 3 − 2 ) cannot be factored any further using integers and the factorization is complete.

Answer: ( x 3 − 2 ) ( x + 2 ) ( x 2 + 2 x + 4 )

Try this! Factor: 9 x 4 + 17 x 2 − 2

Answer: ( 3 x + 1 ) ( 3 x − 1 ) ( x 2 + 2 )

Solving Polynomial Equations by Factoring

In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property A product is equal to zero if and only if at least one of the factors is zero. :

a ⋅ b = 0    if and only if    a = 0  or  b = 0

The zero-product property is true for any number of factors that make up an equation. In other words, if any product is equal to zero, then at least one of the variable factors must be equal to zero. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Solve: 2 x ( x − 4 ) ( 5 x + 3 ) = 0 .

Set each variable factor equal to zero and solve.

2 x = 0   or   x − 4 = 0   or   5 x + 3 = 0 2 x 2 = 0 2 x = 4 5 x 5 = − 3 5 x = 0 x = − 3 5

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

Answer: The solutions are 0, 4, and − 3 5 .

Of course, most equations will not be given in factored form.

Solve: 4 x 3 − x 2 − 100 x + 25 = 0 .

Begin by factoring the left side completely.

4 x 3 − x 2 − 100 x + 25 = 0 F a c t o r   b y   g r o u p i n g . x 2 ( 4 x − 1 ) − 25 ( 4 x − 1 ) = 0 ( 4 x − 1 ) ( x 2 − 25 ) = 0 F a c t o r   a s   a   d i f f e r e n c e   o f   s q u a r e s . ( 4 x − 1 ) ( x + 5 ) ( x − 5 ) = 0

Set each factor equal to zero and solve.

4 x − 1 = 0 or x + 5 = 0 or x − 5 = 0 4 x = 1 x = − 5 x = 5 x = 1 4

Answer: The solutions are 1 4 , −5, and 5.

Using the zero-product property after factoring an equation that is equal to zero is the key to this technique. However, the equation may not be given equal to zero, and so there may be some preliminary steps before factoring. The steps required to solve by factoring The process of solving an equation that is equal to zero by factoring it and then setting each variable factor equal to zero. are outlined in the following example.

Solve: 15 x 2 + 3 x − 8 = 5 x − 7 .

Step 1: Express the equation in standard form, equal to zero. In this example, subtract 5 x from and add 7 to both sides.

15 x 2 + 3 x − 8 = 5 x − 7 15 x 2 − 2 x − 1 = 0

Step 2: Factor the expression.

( 3 x − 1 ) ( 5 x + 1 ) = 0

Step 3: Apply the zero-product property and set each variable factor equal to zero.

3 x − 1 = 0         or         5 x + 1 = 0

Step 4: Solve the resulting linear equations.

3 x − 1 = 0 or 5 x + 1 = 0 3 x = 1 5 x = − 1 x = 1 3 x = − 1 5

Answer: The solutions are 1 3 and − 1 5 . The check is optional.

Solve: ( 3 x + 2 ) ( x + 1 ) = 4 .

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to 4. However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

( 3 x + 2 ) ( x + 1 ) = 4 3 x 2 + 3 x + 2 x + 2 = 4 3 x 2 + 5 x + 2 = 4 3 x 2 + 5 x − 2 = 0

Once it is in standard form, we can factor and then set each factor equal to zero.

                ( 3 x − 1 ) ( x + 2 ) = 0 3 x − 1 = 0               or           x + 2 = 0               3 x = 1                                                 x = − 2                     x = 1 3

Answer: The solutions are 1 3 and −2.

Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0

A root A value in the domain of a function that results in zero. of a function is a value in the domain that results in zero. In other words, the roots occur when the function is equal to zero, f ( x ) = 0 .

Find the roots: f ( x ) = ( x + 2 ) 2 − 4 .

To find roots we set the function equal to zero and solve.

f ( x ) = 0 ( x + 2 ) 2 − 4 = 0 x 2 + 4 x + 4 − 4 = 0 x 2 + 4 x = 0 x ( x + 4 ) = 0

Next, set each factor equal to zero and solve.

x = 0   or     x + 4 = 0 x = − 4

We can show that these x -values are roots by evaluating.

f ( 0 ) = ( 0 + 2 ) 2 − 4         f ( − 4 ) = ( − 4 + 2 ) 2 − 4 = 4 − 4   = ( − 2 ) 2 − 4 = 0         ✓ = 4 − 4 = 0         ✓

Answer: The roots are 0 and −4.

If we graph the function in the previous example we will see that the roots correspond to the x -intercepts of the function. Here the function f is a basic parabola shifted 2 units to the left and 4 units down.

steps in problem solving involving factors of polynomials

Find the roots: f ( x ) = x 4 − 5 x 2 + 4 .

f ( x ) = 0 x 4 − 5 x 2 + 4 = 0 ( x 2 − 1 ) ( x 2 − 4 ) = 0 ( x + 1 ) ( x − 1 ) ( x + 2 ) ( x − 2 ) = 0

x + 1 = 0         or         x − 1 = 0         or         x + 2 = 0         or         x − 2 = 0 x = − 1 x = 1 x = − 2 x = 2

Answer: The roots are −1, 1, −2, and 2.

Graphing the previous function is not within the scope of this course. However, the graph is provided below:

steps in problem solving involving factors of polynomials

Notice that the degree of the polynomial is 4 and we obtained four roots. In general, for any polynomial function with one variable of degree n , the fundamental theorem of algebra Guarantees that there will be as many (or fewer) roots to a polynomial function with one variable as its degree. guarantees n real roots or fewer. We have seen that many polynomials do not factor. This does not imply that functions involving these unfactorable polynomials do not have real roots. In fact, many polynomial functions that do not factor do have real solutions. We will learn how to find these types of roots as we continue in our study of algebra.

Find the roots: f ( x ) = − x 2 + 10 x − 25 .

f ( x ) = 0 − x 2 + 10 x − 25 = 0 − ( x 2 − 10 x + 25 ) = 0 − ( x − 5 ) ( x − 5 ) = 0

Next, set each variable factor equal to zero and solve.

x − 5 = 0 or x − 5 = 0 = 5 x = 5

A solution that is repeated twice is called a double root A root that is repeated twice. . In this case, there is only one solution.

Answer: The root is 5.

The previous example shows that a function of degree 2 can have one root. From the factoring step, we see that the function can be written

f ( x ) = − ( x − 5 ) 2

In this form, we can see a reflection about the x -axis and a shift to the right 5 units. The vertex is the x -intercept, illustrating the fact that there is only one root.

steps in problem solving involving factors of polynomials

Try this! Find the roots of f ( x ) = x 3 + 3 x 2 − x − 3 .

Answer: ±1, −3

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x , where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 40 feet.

We are asked to find the speed x where the safe stopping distance d ( x ) = 40 feet.

d ( x ) = 40 1 20 x 2 + x = 40

To solve for x , rewrite the resulting equation in standard form. In this case, we will first multiply both sides by 20 to clear the fraction.

20 ( 1 20 x 2 + x ) = 20 ( 40 ) x 2 + 20 x = 800 x 2 + 20 x − 800 = 0

Next factor and then set each factor equal to zero.

x 2 + 20 x − 800 = 0 ( x + 40 ) ( x − 20 ) = 0 x + 40 = 0 o r x − 20 = 0 x = − 40 x = 20

The negative answer does not make sense in the context of this problem. Consider x = 20 miles per hour to be the only solution.

Answer: 20 miles per hour

Finding Equations with Given Solutions

We can use the zero-product property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Find a quadratic equation with integer coefficients, given solutions − 3 2 and 1 3 .

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

x = − 3 2 or x = 1 3 2 x = − 3 3 x = 1 2 x + 3 = 0 3 x − 1 = 0

The product of these linear factors is equal to zero when x = − 3 2 or x = 1 3 .

( 2 x + 3 ) ( 3 x − 1 ) = 0

Multiply the binomials and present the equation in standard form.

6 x 2 − 2 x + 9 x − 3 = 0 6 x 2 + 7 x − 3 = 0

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

Answer: 6 x 2 + 7 x − 3 = 0

Find a polynomial function with real roots 1, −2, and 2.

Given solutions to f ( x ) = 0 we can find linear factors.

x = 1 or x = − 2 or x = 2 x − 1 = 0 x + 2 = 0 x − 2 = 0

Apply the zero-product property and multiply.

( x − 1 ) ( x + 2 ) ( x − 2 ) = 0 ( x − 1 ) ( x 2 − 4 ) = 0 x 3 − 4 x − x 2 + 4 = 0 x 3 − x 2 − 4 x + 4 = 0

Answer: f ( x ) = x 3 − x 2 − 4 x + 4

Try this! Find a polynomial equation with integer coefficients, given solutions 1 2 and − 3 4 .

Answer: 8 x 2 + 2 x − 3 = 0

Key Takeaways

  • Factoring and the zero-product property allow us to solve equations.
  • To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting equations are the solutions to the original.
  • Not all polynomial equations can be solved by factoring. We will learn how to solve polynomial equations that do not factor later in the course.
  • A polynomial function can have at most a number of real roots equal to its degree. To find roots of a function, set it equal to zero and solve.
  • To find a polynomial equation with given solutions, perform the process of solving by factoring in reverse.

Topic Exercises

Part a: general factoring.

Factor completely.

50 x 2 − 18

12 x 3 − 3 x

10 x 3 + 65 x 2 − 35 x

15 x 4 + 7 x 3 − 4 x 2

6 a 4 b − 15 a 3 b 2 − 9 a 2 b 3

8 a 3 b − 44 a 2 b 2 + 20 a b 3

36 x 4 − 72 x 3 − 4 x 2 + 8 x

20 x 4 + 60 x 3 − 5 x 2 − 15 x

3 x 5 + 2 x 4 − 12 x 3 − 8 x 2

10 x 5 − 4 x 4 − 90 x 3 + 36 x 2

x 4 − 23 x 2 − 50

2 x 4 − 31 x 2 − 16

− 2 x 5 − 6 x 3 + 8 x

− 36 x 5 + 69 x 3 + 27 x

54 x 5 − 78 x 3 + 24 x

4 x 6 − 65 x 4 + 16 x 2

x 6 − 7 x 3 − 8

x 6 − 25 x 3 − 54

3 x 6 + 4 x 3 + 1

27 x 6 − 28 x 3 + 1

Part B: Solving Polynomial Equations by Factoring

( 6 x − 5 ) ( x + 7 ) = 0

( x + 9 ) ( 3 x − 8 ) = 0

5 x ( 2 x − 5 ) ( 3 x + 1 ) = 0

4 x ( 5 x − 1 ) ( 2 x + 3 ) = 0

( x − 1 ) ( 2 x + 1 ) ( 3 x − 5 ) = 0

( x + 6 ) ( 5 x − 2 ) ( 2 x + 9 ) = 0

( x + 4 ) ( x − 2 ) = 16

( x + 1 ) ( x − 7 ) = 9

( 6 x + 1 ) ( x + 1 ) = 6

( 2 x − 1 ) ( x − 4 ) = 39

x 2 − 15 x + 50 = 0

x 2 + 10 x − 24 = 0

3 x 2 + 2 x − 5 = 0

2 x 2 + 9 x + 7 = 0

1 10 x 2 − 7 15 x − 1 6 = 0

1 4 − 4 9 x 2 = 0

6 x 2 − 5 x − 2 = 30 x + 4

6 x 2 − 9 x + 15 = 20 x − 13

5 x 2 − 23 x + 12 = 4 ( 5 x − 3 )

4 x 2 + 5 x − 5 = 15 ( 3 − 2 x )

( x + 6 ) ( x − 10 ) = 4 ( x − 18 )

( x + 4 ) ( x − 6 ) = 2 ( x + 4 )

4 x 3 − 14 x 2 − 30 x = 0

9 x 3 + 48 x 2 − 36 x = 0

1 3 x 3 − 3 4 x = 0

1 2 x 3 − 1 50 x = 0

− 10 x 3 − 28 x 2 + 48 x = 0

− 2 x 3 + 15 x 2 + 50 x = 0

2 x 3 − x 2 − 72 x + 36 = 0

4 x 3 − 32 x 2 − 9 x + 72 = 0

45 x 3 − 9 x 2 − 5 x + 1 = 0

x 3 − 3 x 2 − x + 3 = 0

x 4 − 5 x 2 + 4 = 0

4 x 4 − 37 x 2 + 9 = 0

Find the roots of the given functions.

f ( x ) = x 2 + 10 x − 24

f ( x ) = x 2 − 14 x + 48

f ( x ) = − 2 x 2 + 7 x + 4

f ( x ) = − 3 x 2 + 14 x + 5

f ( x ) = 16 x 2 − 40 x + 25

f ( x ) = 9 x 2 − 12 x + 4

g ( x ) = 8 x 2 + 3 x

g ( x ) = 5 x 2 − 30 x

p ( x ) = 64 x 2 − 1

q ( x ) = 4 x 2 − 121

f ( x ) = 1 5 x 3 − 1 x 2 − 1 20 x + 1 4

f ( x ) = 1 3 x 3 + 1 2 x 2 − 4 3 x − 2

g ( x ) = x 4 − 13 x 2 + 36

g ( x ) = 4 x 4 − 13 x 2 + 9

f ( x ) = ( x + 5 ) 2 − 1

g ( x ) = − ( x + 5 ) 2 + 9

f ( x ) = − ( 3 x − 5 ) 2

g ( x ) = − ( x + 2 ) 2 + 4

Given the graph of a function, determine the real roots.

steps in problem solving involving factors of polynomials

The sides of a square measure x − 2 units. If the area is 36 square units, then find x .

The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

The profit in dollars generated by producing and selling n bicycles per week is given by the formula P ( n ) = − 5 n 2 + 400 n − 6000 . How many bicycles must be produced and sold to break even?

The height in feet of an object dropped from the top of a 64-foot building is given by h ( t ) = − 16 t 2 + 64 where t represents the time in seconds after it is dropped. How long will it take to hit the ground?

A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

steps in problem solving involving factors of polynomials

What is the length of each side of the cardboard sheet if the volume of the box is to be 98 cubic inches?

The height of a triangle is 4 centimeters less than twice the length of its base. If the total area of the triangle is 48 square centimeters, then find the lengths of the base and height.

A uniform border is to be placed around an 8 × 10 inch picture.

steps in problem solving involving factors of polynomials

If the total area including the border must be 168 square inches, then how wide should the border be?

The area of a picture frame including a 3-inch wide border is 120 square inches.

steps in problem solving involving factors of polynomials

If the width of the inner area is 2 inches less than its length, then find the dimensions of the inner area.

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 75 feet.

A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula R ( n ) = 12 n − 0.6 n 2 where n represents the number of palettes of product sold ( 0 ≤ n < 20 ) . Determine the number of palettes sold in a day if the revenue was 45 thousand dollars.

Part C: Finding Equations with Given Solutions

Find a polynomial equation with the given solutions.

Find a function with the given roots.

2 5 , − 1 3

5 double root

−3 double root

Recall that if | X | = p , then X = − p or X = p . Use this to solve the following absolute value equations.

| x 2 − 8 | = 8

| 2 x 2 − 9 | = 9

| x 2 − 2 x − 1 | = 2

| x 2 − 8 x + 14 | = 2

| 2 x 2 − 4 x − 7 | = 9

| x 2 − 3 x − 9 | = 9

Part D: Discussion Board

Explain to a beginning algebra student the difference between an equation and an expression.

What is the difference between a root and an x -intercept? Explain.

Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.

Research and discuss the fundamental theorem of algebra.

2 ( 5 x + 3 ) ( 5 x − 3 )

5 x ( x + 7 ) ( 2 x − 1 )

3 a 2 b ( 2 a + b ) ( a − 3 b )

4 x ( x − 2 ) ( 3 x + 1 ) ( 3 x − 1 )

x 2 ( 3 x + 2 ) ( x + 2 ) ( x − 2 )

( x 2 + 2 ) ( x + 5 ) ( x − 5 )

  • − 2 x ( x 2 + 4 ) ( x − 1 ) ( x + 1 )

6 x ( x + 1 ) ( x − 1 ) ( 3 x + 2 ) ( 3 x − 2 )

( x + 1 ) ( x 2 − x + 1 ) ( x − 2 ) ( x 2 + 2 x + 4 )

( 3 x 3 + 1 ) ( x + 1 ) ( x 2 − x + 1 )

0, 5 2 , − 1 3

− 1 2 , 1, 5 3

− 5 3 , 1 2

0, − 3 2 , 5

± 1 3 , 1 5

−3, −1, 0, 2

20 or 60 bicycles

30 miles per hour

x 2 − 2 x − 15 = 0

3 x 2 − 7 x + 2 = 0

x 2 + 4 x = 0

x 2 − 49 = 0

x 3 − x 2 − 9 x + 9 = 0

f ( x ) = 6 x 2 − 7 x + 2

f ( x ) = 16 x 2 − 9

f ( x ) = x 2 − 10 x + 25

f ( x ) = x 3 − 2 x 2 − 3 x

Answer may vary

  • 6.4 General Strategy for Factoring Polynomials
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.1 Solve Systems of Linear Equations with Two Variables
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Recognize and use the appropriate method to factor a polynomial completely

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

You have now become acquainted with all the methods of factoring that you will need in this course. The following chart summarizes all the factoring methods we have covered, and outlines a strategy you should use when factoring polynomials.

General Strategy for Factoring Polynomials

Use a general strategy for factoring polynomials..

  • Step 1. Is there a greatest common factor? Factor it out.
  • Is it a sum? Of squares? Sums of squares do not factor. Of cubes? Use the sum of cubes pattern.
  • Is it a difference? Of squares? Factor as the product of conjugates. Of cubes? Use the difference of cubes pattern.
  • Is it of the form x 2 + b x + c ? x 2 + b x + c ? Undo FOIL.
  • Is it of the form a x 2 + b x + c ? a x 2 + b x + c ? If a and c are squares, check if it fits the trinomial square pattern. Use the trial and error or “ac” method.
  • Use the grouping method.
  • Step 3. Check. Is it factored completely? Do the factors multiply back to the original polynomial?

Remember, a polynomial is completely factored if, other than monomials, its factors are prime !

Example 6.35

Factor completely: 7 x 3 − 21 x 2 − 70 x . 7 x 3 − 21 x 2 − 70 x .

Try It 6.69

Factor completely: 8 y 3 + 16 y 2 − 24 y . 8 y 3 + 16 y 2 − 24 y .

Try It 6.70

Factor completely: 5 y 3 − 15 y 2 − 270 y . 5 y 3 − 15 y 2 − 270 y .

Be careful when you are asked to factor a binomial as there are several options!

Example 6.36

Factor completely: 24 y 2 − 150 . 24 y 2 − 150 .

Try It 6.71

Factor completely: 16 x 3 − 36 x . 16 x 3 − 36 x .

Try It 6.72

Factor completely: 27 y 2 − 48 . 27 y 2 − 48 .

The next example can be factored using several methods. Recognizing the trinomial squares pattern will make your work easier.

Example 6.37

Factor completely: 4 a 2 − 12 a b + 9 b 2 . 4 a 2 − 12 a b + 9 b 2 .

Try It 6.73

Factor completely: 4 x 2 + 20 x y + 25 y 2 . 4 x 2 + 20 x y + 25 y 2 .

Try It 6.74

Factor completely: 9 x 2 − 24 x y + 16 y 2 . 9 x 2 − 24 x y + 16 y 2 .

Remember, sums of squares do not factor, but sums of cubes do!

Example 6.38

Factor completely 12 x 3 y 2 + 75 x y 2 . 12 x 3 y 2 + 75 x y 2 .

Try It 6.75

Factor completely: 50 x 3 y + 72 x y . 50 x 3 y + 72 x y .

Try It 6.76

Factor completely: 27 x y 3 + 48 x y . 27 x y 3 + 48 x y .

When using the sum or difference of cubes pattern, being careful with the signs.

Example 6.39

Factor completely: 24 x 3 + 81 y 3 . 24 x 3 + 81 y 3 .

Try It 6.77

Factor completely: 250 m 3 + 432 n 3 . 250 m 3 + 432 n 3 .

Try It 6.78

Factor completely: 2 p 3 + 54 q 3 . 2 p 3 + 54 q 3 .

Example 6.40

Factor completely: 3 x 5 y − 48 x y . 3 x 5 y − 48 x y .

Try It 6.79

Factor completely: 4 a 5 b − 64 a b . 4 a 5 b − 64 a b .

Try It 6.80

Factor completely: 7 x y 5 − 7 x y . 7 x y 5 − 7 x y .

Example 6.41

Factor completely: 4 x 2 + 8 b x − 4 a x − 8 a b . 4 x 2 + 8 b x − 4 a x − 8 a b .

Try It 6.81

Factor completely: 6 x 2 − 12 x c + 6 b x − 12 b c . 6 x 2 − 12 x c + 6 b x − 12 b c .

Try It 6.82

Factor completely: 16 x 2 + 24 x y − 4 x − 6 y . 16 x 2 + 24 x y − 4 x − 6 y .

Taking out the complete GCF in the first step will always make your work easier.

Example 6.42

Factor completely: 40 x 2 y + 44 x y − 24 y . 40 x 2 y + 44 x y − 24 y .

Try It 6.83

Factor completely: 4 p 2 q − 16 p q + 12 q . 4 p 2 q − 16 p q + 12 q .

Try It 6.84

Factor completely: 6 p q 2 − 9 p q − 6 p . 6 p q 2 − 9 p q − 6 p .

When we have factored a polynomial with four terms, most often we separated it into two groups of two terms. Remember that we can also separate it into a trinomial and then one term.

Example 6.43

Factor completely: 9 x 2 − 12 x y + 4 y 2 − 49 . 9 x 2 − 12 x y + 4 y 2 − 49 .

Try It 6.85

Factor completely: 4 x 2 − 12 x y + 9 y 2 − 25 . 4 x 2 − 12 x y + 9 y 2 − 25 .

Try It 6.86

Factor completely: 16 x 2 − 24 x y + 9 y 2 − 64 . 16 x 2 − 24 x y + 9 y 2 − 64 .

Practice Makes Perfect

In the following exercises, factor completely.

2 n 2 + 13 n − 7 2 n 2 + 13 n − 7

8 x 2 − 9 x − 3 8 x 2 − 9 x − 3

a 5 + 9 a 3 a 5 + 9 a 3

75 m 3 + 12 m 75 m 3 + 12 m

121 r 2 − s 2 121 r 2 − s 2

49 b 2 − 36 a 2 49 b 2 − 36 a 2

8 m 2 − 32 8 m 2 − 32

36 q 2 − 100 36 q 2 − 100

25 w 2 − 60 w + 36 25 w 2 − 60 w + 36

49 b 2 − 112 b + 64 49 b 2 − 112 b + 64

m 2 + 14 m n + 49 n 2 m 2 + 14 m n + 49 n 2

64 x 2 + 16 x y + y 2 64 x 2 + 16 x y + y 2

7 b 2 + 7 b − 42 7 b 2 + 7 b − 42

30 n 2 + 30 n + 72 30 n 2 + 30 n + 72

3 x 4 y − 81 x y 3 x 4 y − 81 x y

4 x 5 y − 32 x 2 y 4 x 5 y − 32 x 2 y

k 4 − 16 k 4 − 16

m 4 − 81 m 4 − 81

5 x 5 y 2 − 80 x y 2 5 x 5 y 2 − 80 x y 2

48 x 5 y 2 − 243 x y 2 48 x 5 y 2 − 243 x y 2

15 p q − 15 p + 12 q − 12 15 p q − 15 p + 12 q − 12

12 a b − 6 a + 10 b − 5 12 a b − 6 a + 10 b − 5

4 x 2 + 40 x + 84 4 x 2 + 40 x + 84

5 q 2 − 15 q − 90 5 q 2 − 15 q − 90

4 u 5 + 4 u 2 v 3 4 u 5 + 4 u 2 v 3

5 m 4 n + 320 m n 4 5 m 4 n + 320 m n 4

4 c 2 + 20 c d + 81 d 2 4 c 2 + 20 c d + 81 d 2

25 x 2 + 35 x y + 49 y 2 25 x 2 + 35 x y + 49 y 2

10 m 4 − 6250 10 m 4 − 6250

3 v 4 − 768 3 v 4 − 768

36 x 2 y + 15 x y − 6 y 36 x 2 y + 15 x y − 6 y

60 x 2 y − 75 x y + 30 y 60 x 2 y − 75 x y + 30 y

8 x 3 − 27 y 3 8 x 3 − 27 y 3

64 x 3 + 125 y 3 64 x 3 + 125 y 3

y 6 − 1 y 6 − 1

y 6 + 1 y 6 + 1

9 x 2 − 6 x y + y 2 − 49 9 x 2 − 6 x y + y 2 − 49

16 x 2 − 24 x y + 9 y 2 − 64 16 x 2 − 24 x y + 9 y 2 − 64

( 3 x + 1 ) 2 − 6 ( 3 x + 1 ) + 9 ( 3 x + 1 ) 2 − 6 ( 3 x + 1 ) + 9

( 4 x − 5 ) 2 − 7 ( 4 x − 5 ) + 12 ( 4 x − 5 ) 2 − 7 ( 4 x − 5 ) + 12

Writing Exercises

Explain what it mean to factor a polynomial completely.

The difference of squares y 4 − 625 y 4 − 625 can be factored as ( y 2 − 25 ) ( y 2 + 25 ) . ( y 2 − 25 ) ( y 2 + 25 ) . But it is not completely factored. What more must be done to completely factor.

Of all the factoring methods covered in this chapter (GCF, grouping, undo FOIL, ‘ac’ method, special products) which is the easiest for you? Which is the hardest? Explain your answers.

Create three factoring problems that would be good test questions to measure your knowledge of factoring. Show the solutions.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
  • Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/6-4-general-strategy-for-factoring-polynomials

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How to Solve Polynomials

Last Updated: January 22, 2024 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 12 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 334,512 times.

A polynomial is an expression made up of adding and subtracting terms. A terms can consist of constants, coefficients, and variables. When solving polynomials, you usually trying to figure out for which x-values y=0. Lower-degree polynomials will have zero, one or two real solutions, depending on whether they are linear polynomials or quadratic polynomials. These types of polynomials can be easily solved using basic algebra and factoring methods. For help solving polynomials of a higher degree, read Solve Higher Degree Polynomials .

Solving a Linear Polynomial

Step 1 Determine whether you have a linear polynomial.

Solving a Quadratic Polynomial

Step 1 Determine whether you have a quadratic polynomial.

Community Q&A

Donagan

  • Remember the order of operations while you work -- First work in the parenthesis, then do the multiplication and division, and finally do the addition and subtraction. [17] X Research source Thanks Helpful 0 Not Helpful 0
  • Don't fret if you get different variables, like t, or if you see an equation set to f(x) instead of 0. If the question wants roots, zeros, or factors, just treat it like any other problem. Thanks Helpful 3 Not Helpful 3

steps in problem solving involving factors of polynomials

You Might Also Like

Differentiate Polynomials

  • ↑ https://www.cuemath.com/algebra/linear-polynomial/
  • ↑ https://www.math.utah.edu/~wortman/1050-text-calp.pdf
  • ↑ https://www.mathsisfun.com/algebra/polynomials-solving.html
  • ↑ David Jia. Academic Tutor. Expert Interview. 7 January 2021.
  • ↑ http://www.mathwords.com/c/constant.htm
  • ↑ https://www.cuemath.com/algebra/factorization-of-quadratic-polynomials/
  • ↑ http://www.themathpage.com/aprecalc/quadratic-equation.htm#double
  • ↑ https://www.math.utah.edu/~wortman/1050-text-qp.pdf
  • ↑ https://www.khanacademy.org/math/algebra/quadratics/solving-quadratic-equations-by-factoring/v/example-1-solving-a-quadratic-equation-by-factoring
  • ↑ https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-grouping/a/factoring-by-grouping
  • ↑ https://content.byui.edu/file/b8b83119-9acc-4a7b-bc84-efacf9043998/1/Math-1-6-1.html

About This Article

David Jia

To solve a linear polynomial, set the equation to equal zero, then isolate and solve for the variable. A linear polynomial will have only one answer. If you need to solve a quadratic polynomial, write the equation in order of the highest degree to the lowest, then set the equation to equal zero. Rewrite the expression as a 4-term expression and factor the equation by grouping. Rewrite the polynomial as 2 binomials and solve each one. If you want to learn how to simplify and solve your terms in a polynomial equation, keep reading the article! Did this summary help you? Yes No

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Mathematics LibreTexts

6.6: Solving Equations by Factoring

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  • Page ID 23743

Learning Objectives

  • Verify solutions to quadratic equations.
  • Solve quadratic equations by factoring.
  • Determine a quadratic equation with given solutions.
  • Solve polynomial equations by factoring.

Solving Quadratic Equations by Factoring

Learning how to solve equations is one of our main goals in algebra. Up to this point, we have solved linear equations, which are of degree 1. In this section, we will learn a technique that can be used to solve certain equations of degree 2. A quadratic equation is any equation that can be written in the standard form

\[ax^{2}+bx+c=0\],

where \(a, b\), and \(c\) are real numbers and \(a≠0\). The following are some examples of quadratic equations, all of which will be solved in this section:

\(x^{2}+x-6=0\)

\(4x^{2}-9=0\)

\(2x^{2}+10x+20=-3x+5\)

A solution of a quadratic equation in standard form is called a root. Quadratic equations can have two real solutions, one real solution, or no real solution. The quadratic equation \(x^{2}+x−6=0\) has two solutions, namely, \(x=−3\) and \(x=2\).

Example \(\PageIndex{1}\)

Verify that \(x=−3\) and \(x=2\) are solutions to \(x^{2}+x−6=0\).

To verify solutions, substitute the values for \(x\) and then simplify to see if a true statement results.

\(\begin{array} {r|r} {\color{Cerulean}{Check:}\color{black}{\:x=-3}}&{\color{Cerulean}{Check:}\color{black}{\:x=2}} \\ {x^{2}+x-6=0}&{x^{2}+x-6=0}\\{(\color{OliveGreen}{-3}\color{black}{)^{2}+(}\color{OliveGreen}{-3}\color{black}{)-6=0}}&{(\color{OliveGreen}{2}\color{black}{)^{2}+(}\color{OliveGreen}{2}\color{black}{)-6=0}}\\{9-3-6=0}&{4+2-6=0}\\{0=0\quad\color{Cerulean}{\checkmark}}&{0=0\quad\color{Cerulean}{\checkmark}} \end{array}\)

Both values produce true statements. Therefore, they are both solutions to the equation.

Our goal is to develop algebraic techniques for finding solutions to quadratic equations. The first technique requires the zero-product property:

If \(a\cdot b=0\), then \(a=0\) or \(b=0\)

In other words, if any product is equal to zero, then one or both of the variable factors must be equal to zero.

Example \(\PageIndex{2}\)

\((x−8)(x+7)=0\).

This equation consists of a product of two quantities equal to zero; therefore, the zero-product property applies. One or both of the quantities must be zero.

\(\begin{array} {ccc} {(x-8)=0}&{\text{or}}&{(x+7)=0}\\{x-8\color{Cerulean}{+8}\color{black}{=0}\color{Cerulean}{+8}}&{}&{x+7\color{Cerulean}{-7}\color{black}{=0}\color{Cerulean}{-7}}\\{x=8}&{}&{x=-7} \end{array}\)

To verify that these are solutions, substitute them for the variable \(x\).

\(\begin{array}{r|r} {\color{Cerulean}{Check:}\color{black}{\:x=8}}&{\color{Cerulean}{Check:}\color{black}{\:x=-7}}\\{(x-8)(x+7)=0}&{(x-8)(x+7)=0}\\{(8-8)(8+7)=0}&{(-7+8)(-7+7)=0}\\{(0)(15)=0}&{(1)(0)=0}\\{0=0\quad\color{Cerulean}{\checkmark}}&{0=0\quad\color{Cerulean}{\checkmark}} \end{array}\)

Notice that each solution produces a factor that is equal to zero.

The solutions are \(8\) and \(−7\).

The quadratic equation may not be given in its factored form.

Example \(\PageIndex{3}\)

\(x^{2}+3x−10=0\).

The goal is to produce a product that is equal to zero. We can do that by factoring the trinomial on the left side of the equation.

\(\begin{array}{cc}{x^{2}+3x-10=0}&{}\\{(x\quad\color{Cerulean}{?}\color{black}{)(x\quad}\color{Cerulean}{?}\color{black}{)=0}}&{\color{Cerulean}{-10=5(-2)}}\\{}&{\color{Cerulean}{and\: 3=5+(-2)}}\\{(x+5)(x-2)=0}&{} \end{array}\)

Next, apply the zero-product property and set each factor equal to zero.

\(x+5=0 \quad\text{or}\quad x-2=0\)

This leaves us with two linear equations, each of which can be solved for \(x\).

\(\begin{array}{cc} {x+5\color{Cerulean}{-5}\color{black}{=0}\color{Cerulean}{-5}}&{x-2\color{Cerulean}{+2}\color{black}{=0}\color{Cerulean}{+2}}\\{x=-5}&{x=2} \end{array}\)

Check the solutions by substituting into the original equation to verify that we obtain true statements.

\(\begin{array}{r|r}{\color{Cerulean}{Check:}\color{black}{\:x=-5}}&{\color{Cerulean}{Check:}\color{black}{\:x=2}}\\{x^{2}+3x-10=0}&{x^{2}+3x-10=0}\\{(\color{OliveGreen}{-5}\color{black}{)^{2}+3(}\color{OliveGreen}{-5}\color{black}{)-10=0}}&{(\color{OliveGreen}{2}\color{black}{)^{2}+3(}\color{OliveGreen}{2}\color{black}{)-10=0}}\\{25-15-10=0}&{4+6-10=0}\\{0=0\quad\color{Cerulean}{\checkmark}}&{0=0\quad\color{Cerulean}{\checkmark}} \end{array}\)

The solutions are \(-5\) and \(2\).

Using the zero-product property after factoring a quadratic equation in standard form is the key to this technique. However, the quadratic equation may not be given in standard form, and so there may be some preliminary steps before factoring. The steps required to solve by factoring are outlined in the following example.

Example \(\PageIndex{4}\)

\(2x^{2}+10x+20=−3x+5\).

Step 1 : Express the quadratic equation in standard form. For the zero-product property to apply, the quadratic expression must be equal to zero. Use the addition and subtraction properties of equality to combine opposite-side like terms and obtain zero on one side of the equation. In this example, add \(3x\) to and subtract \(5\) from both sides.

\(\begin{aligned} 2x^{2}+10x+20\color{Cerulean}{+3x}&\color{black}{=-3x+5}\color{Cerulean}{+3x} \\ 2x^{2}+13x+20&=5 \\ 2x^{2}+13x+20\color{Cerulean}{-5}&\color{black}{=5}\color{Cerulean}{-5} \\ 2x^{2}+13x+15&=0 \end{aligned}\)

Step 2 : Factor the quadratic expression.

\((2x+3)(x+5)=0\)

Step 3 : Apply the zero-product property and set each variable factor equal to zero.

\(2x+3=0\quad\text{or}\quad x+5=0\)

Step 4 : Solve the resulting linear equations.

\(\begin{array} {ccc} {2x+3=0}&{\text{or}}&{x+5=0}\\{2x=-3}&{}&{x=-5}\\{\frac{2x}{\color{Cerulean}{2}}\color{black}{=\frac{-3}{\color{Cerulean}{2}}}}&{}&{}\\{x=-\frac{3}{2}}&{}&{} \end{array}\)

The solutions are \(-5\) and \(-\frac{3}{2}\). The check is optional.

Example \(\PageIndex{5}\)

\(9x^{2}+1=6x\).

Write this in standard form by subtracting \(6x\ from both sides.

\(\begin{aligned} 9x^{2}+1\color{Cerulean}{-6x}&\color{black}{=6x}\color{Cerulean}{-6x} \\ 9x^{2}-6x+1&=0 \end{aligned}\)

Once the equation is in standard form, equal to zero, factor.

\((3x-1)(3x-1)=0\)

This is a perfect square trinomial. Hence setting each factor equal to zero results in a repeated solution.

\(\begin{array}{ccc}{3x-1=0}&{\text{or}}&{3x-1=0}\\{3x=1}&{}&{3x=1}\\{x=\frac{1}{3}}&{}&{x=\frac{1}{3}}\\ \end{array}\)

A repeated solution is called a double root and does not have to be written twice.

The solution is \(\frac{1}{3}\).

Exercise \(\PageIndex{1}\)

\(x^{2}−3x=28\).

\(x=−4\) or \(x=7\)

Not all quadratic equations in standard form are trinomials. We often encounter binomials.

Example \(\PageIndex{6}\)

\(x^{2}−9=0\).

This quadratic equation is given in standard form, where the binomial on the left side is a difference of squares. Factor as follows:

\(\begin{aligned} x^{2}-9&=0 \\ (x+3)(x-3)&=0 \end{aligned}\)

Next, set each factor equal to zero and solve.

\(\begin{array}{ccc}{x+3=0}&{\text{or}}&{x-3=0}\\{x=-3}&{}&{x=3}\end{array}\)

The solutions are \(3\) and \(−3\), which can also be written as \(±3\).

Example \(\PageIndex{7}\)

\(5x^{2}=15x\)

By inspection, we see that \(x=0\) is a solution to this quadratic equation. Since dividing by zero is undefined, we want to avoid dividing both sides of this equation by \(x\). In general, we wish to avoid dividing both sides of any equation by a variable or an expression containing a variable. We will discuss this in more detail later. The first step is to rewrite this equation in standard form with zero on one side.

\(\begin{aligned} 5x^{2}&=15x \\ 5x^{2}\color{Cerulean}{-15x}&\color{black}{=15x}\color{Cerulean}{-15x\quad Subtract\:15x\:from\:both\:sides.} \\ 5x^{2}-15x&=0 \end{aligned}\)

Next, factor the expression. Notice that the binomial on the left has a GCF of \(5x\).

\(\begin{aligned} 5x^{2}-15x&=0\\5x(x-3)&=0 \end{aligned}\)

Set each factor to equal to zero.

\(\begin{array}{ccc}{5x=0}&{\text{or}}&{x-3=0}\\{\frac{5x}{\color{Cerulean}{5}}\color{black}{=\frac{0}{\color{Cerulean}{5}}}}&{}&{x=3}\\{x=0}&{}&{} \end{array}\)

The solutions are \(0\) and \(3\).

Example \(\PageIndex{8}\)

\((2x+1)(x+5)=11\).

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to \(11\). However, this would lead to incorrect results. We must rewrite the equation in standard form, equal to zero, so that we can apply the zero-product property.

\(\begin{aligned}(2x+1)(x+5)&=11 \\ 2x^{2}+10x+x+5&=11 \\ 2x^{2}+11x+5\color{Cerulean}{-11}&\color{black}{=11}\color{Cerulean}{-11}\\2x^{2}+11x-6&=0 \end{aligned}\)

Once it is in standard form, we can factor and then set each factor equal to zero.

\((2x-1)(x+6)=0\)

\(\begin{array}{ccc}{2x-1=0}&{\text{or}}&{x+6=0}\\{2x=1}&{}&{x=-6}\\{x=\frac{1}{2}}&{}&{} \end{array}\)

The solutions are \(\frac{1}{2}\) and \(-6\).

Example \(\PageIndex{9}\)

\(15x^{2}−25x+10=0\).

We begin by factoring out the GCF of \(5\). Then factor the resulting trinomial.

\(\begin{aligned} 15x^{2}-25x+10&=0 \\ 5(3x^{2}-5x+2)&=0\\5(3x-2)(x-1)&=0 \end{aligned}\)

Next, we set each variable factor equal to zero and solve for \(x\).

\(\begin{array}{ccc} {3x-2=0}&{\text{or}}&{x-1=0}\\{3x=2}&{}&{x=1}\\{x=\frac{2}{3}}&{}&{} \end{array}\)

Notice that the factor \(5\) is not a variable factor and thus did not contribute to the solution set.

The solutions are \(\frac{2}{3}\) and \(1\).

Example \(\PageIndex{10}\)

\(52x^{2}+76x−13=0\).

Clear the fractions by multiplying both sides of the equation by the LCD, which is equal to \(6\).

\(\begin{aligned} \frac{5}{2}x^{2}+\frac{7}{6}x-\frac{1}{3}&=0 \\ \color{Cerulean}{6}\color{black}{\cdot} \left( \frac{5}{2}x^{2}+\frac{7}{6}x-\frac{1}{3} \right) &=\color{Cerulean}{6}\color{black}{\cdot (0)} \\ 15x^{2}+7x-2&=0 \end{aligned}\)

At this point, we have an equivalent equation with integer coefficients and can factor as usual. Begin with the factors of \(15\) and \(2\).

\(\begin{array}{cc}{15=1\cdot 15}&{2=1\cdot 2} \\ {=3\cdot 5}&{} \end{array}\)

The coefficient of the middle term is \(7=3(−1)+5(2)\). Factor as follows:

\(\begin{aligned} 15x^{2}+7x-2&=0 \\ (3x+2)(5x-1)&=0 \end{aligned}\)

Set each factor equal to zero and solve.

\(\begin{array}{ccc}{3x+2=0}&{\text{or}}&{5x-1=0}\\{3x=-2}&{}&{5x=1}\\{x=-\frac{2}{3}}&{}&{x=\frac{1}{5}} \end{array}\)

The solutions are \(-\frac{2}{3}\) and \(\frac{1}{5}\).

Exercise \(\PageIndex{2}\)

\(4x^{2}-9=0\).

\(-\frac{3}{2}\) and \(\frac{3}{2}\)

Finding Equations with Given Solutions

The zero-product property states,

And, in fact, the converse is true as well:

If \(a=0\) or \(b=0\), then \(ab=0\)

When this is the case, we can write the following:

\(a\cdot b =0\), if and only if \(a=0\) or \(b=0\)

We use this property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Example \(\PageIndex{11}\)

Find a quadratic equation with solutions \(−7\) and \(2\).

Given the solutions, we can determine two linear factors.

\(\begin{array} {ccc}{x=-7}&{\text{or}}&{x=2}\\{x+7=0}&{}&{x-2=0} \end{array}\)

The product of these linear factors is equal to zero when \(x=-7\) or \(x=2\):

\((x+7)(x-2)=0\)

Multiply the binomials and present the equation in standard form.

\(\begin{aligned} x^{2}-2x+7x-14&=0 \\ x^{2}+5x-14&=0 \end{aligned}\)

\(x^{2}+5x-14=0\).

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found aboveis not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

Example \(\PageIndex{12}\)

Find a quadratic equation with integer coefficients, given solutions \(\frac{1}{2}\) and \(−\frac{3}{4}\).

To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

\(\begin{array} {ccc}{x=\frac{1}{2}}&{\text{or}}&{x=-\frac{3}{4}}\\{\color{Cerulean}{2}\color{black}{\cdot x =}\color{Cerulean}{2}\color{black}{\cdot \frac{1}{2}}}&{}&{\color{Cerulean}{4}\color{black}{\cdot x = -\frac{3}{4}\cdot}\color{Cerulean}{4}}\\{2x=1}&{}&{4x=-3}\\{2x-1=0}&{}&{4x+3=0} \end{array}\)

Apply the zero-product property and multiply.

\(\begin{aligned} (2x-1)(4x+3)&=0 \\ 8x^{2}+6x-4x-3&=0 \\ 8x^{2}+2x-3&=0 \end{aligned}\)

\(8x^{2}+2x-3=0\)

Exercise \(\PageIndex{3}\)

Find a quadratic equation with integer coefficients, given solutions \(−1\) and \(\frac{2}{3}\).

\(3x^{2}+x-2=0\)

Solving Polynomial Equations by Factoring

The zero-product property is true for any number of factors that make up an equation. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Example \(\PageIndex{13}\)

\(3x(x−5)(3x−2)=0\).

Set each variable factor equal to zero and solve.

\(\begin{array}{ccccc} {3x=0}&{\text{or}}&{x-5=0}&{\text{or}}&{3x-2=0}\\{\frac{3x}{\color{Cerulean}{3}}\color{black}{=\frac{0}{\color{Cerulean}{3}}}}&{}&{x=5}&{}&{\frac{3x}{\color{Cerulean}{3}}\color{black}{=\frac{2}{\color{Cerulean}{3}}}}\\{x=0}&{}&{}&{}&{x=\frac{2}{3}}\end{array}\)

The solutions are \(0, 5\), and \(\frac{2}{3}\).

Of course, we cannot expect the equation to be given in factored form.

Example \(\PageIndex{14}\)

\(x^{3}+2x^{2}−9x−18=0\).

Begin by factoring the left side completely.

\(\begin{array}{rl} {x^{3}+2x^{2}-9x-18=0}&{\color{Cerulean}{Factor\:by\:grouping.}}\\{x^{2}(x+2)-9(x+2)=0}&{}\\{(x+2)(x^{2}-9)=0}&{\color{Cerulean}{Factor\:as\:a\:difference\:of\:squares.}}\\{(x+2)(x+3)(x-3)=0}&{} \end{array}\)

\(\begin{array} {ccccc}{x+2=0}&{\text{or}}&{x+3=0}&{\text{or}}&{x-3=0}\\{x=-2}&{}&{x=-3}&{}&{x=3} \end{array}\)

The solutions are \(-2, -3\), and \(3\).

Notice that the degree of the polynomial is \(3\) and we obtained three solutions. In general, for any polynomial equation with one variable of degree \(n\), the fundamental theorem of algebra guarantees \(n\) real solutions or fewer. We have seen that many polynomials do not factor. This does not imply that equations involving these unfactorable polynomials do not have real solutions. In fact, many polynomial equations that do not factor do have real solutions. We will learn how to solve these types of equations as we continue in our study of algebra.

Exercise \(\PageIndex{4}\)

\(−10x^{3}−18x^{2}+4x=0\).

\(−2, 0, \frac{1}{5}\)

Key Takeaways

  • A polynomial can have at most a number of solutions equal to its degree. Therefore, quadratic equations can have up to two real solutions.
  • To solve a quadratic equation, first write it in standard form. Once the quadratic expression is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting linear equations are the solutions to the quadratic equation.
  • Not all quadratic equations can be solved by factoring. We will learn how to solve quadratic equations that do not factor later in the course.
  • To find a quadratic equation with given solutions, perform the process of solving by factoring in reverse.
  • If any polynomial is factored into linear factors and is set to zero, then we can determine the solutions by setting each variable factor equal to zero and solving each individually.

Exercise \(\PageIndex{5}\) Solutions to Quadratic Equations

Determine whether the given set of values are solutions to the quadratic equation.

  • \({−3, 5}; x^{2}−2x−15=0\)
  • \({7, −1}; x^{2}−6x−7=0\)
  • \({−\frac{1}{2}, \frac{1}{2} }; x^{2}−14=0\)
  • \({−\frac{3}{4}, \frac{3}{4} }; x^{2}−916=0\)
  • \({−3, 2}; x^{2}−x−6=0\)
  • \({−5, 1}; x^{2}−4x−5=0\)

Exercise \(\PageIndex{6}\) Solutions to Quadratic Equations

  • \((x−3)(x+2)=0\)
  • \((x+5)(x+1)=0\)
  • \((2x−1)(x−4)=0\)
  • \((3x+1)(3x−1)=0\)
  • \((x−2)^{2}=0\)
  • \((5x+3)^{2}=0\)
  • \(7x(x−5)=0\)
  • \(-2x(2x−3)=0\)
  • \((x−12)(x+34)=0\)
  • \((x+58)(x−38)=0\)
  • \((14x+12)(16x−23)=0\)
  • \((15x−3)^{2}=0\)
  • \(−5(x+1)(x−2)=0\)
  • \(12(x−7)(x−6)=0\)
  • \((x+5)(x−1)=0\)
  • \(−2(3x−2)(2x+5)=0\)
  • \(5(7x−8)^{2}=0\)

1. \(−2, 3\)

3. \(\frac{1}{2}, 4\)

7. \(0, 5\)

9. \(−\frac{3}{4}, \frac{1}{2}\)

11. \(−2, 4\)

13. \(−1, 2\)

15. \(−5, 1\)

17. \(−\frac{5}{2}, \frac{2}{3}\)

Exercise \(\PageIndex{7}\) Solve by Factoring

  • \(x^{2}−x−6=0\)
  • \(x^{2}+3x−10=0\)
  • \(y^{2}−10y+24=0\)
  • \(y^{2}+6y−27=0\)
  • \(x^{2}−14x+40=0\)
  • \(x^{2}+14x+49=0\)
  • \(x^{2}−10x+25=0\)
  • \(3x^{2}+2x−1=0\)
  • \(5x^{2}−9x−2=0\)
  • \(7y^{2}+20y−3=0\)
  • \(9x^{2}−42x+49=0\)
  • \(25x^{2}+30x+9=0\)
  • \(2y^{2}+y−3=0\)
  • \(7x^{2}−11x−6=0\)
  • \(2x^{2}=−15x+8\)
  • \(8x−5=3x^{2}\)
  • \(x^{2}−36=0\)
  • \(x^{2}−100=0\)
  • \(4x^{2}−81=0\)
  • \(49x^{2}−4=0\)
  • \(x^{2}=4\)
  • \(9y^{2}=1\)
  • \(16y^{2}=25\)
  • \(36x^{2}=25\)
  • \(4x^{2}−36=0\)
  • \(2x^{2}−18=0\)
  • \(10x^{2}+20x=0\)
  • \(−3x^{2}+6x=0\)
  • \(25x^{2}=50x\)
  • \(x^{2}=0\)
  • \((x+1)^{2}−25=0\)
  • \((x−2)^{2}−36=0\)
  • \(5x(x−4)=−4+x\)
  • \((x−1)(x−10)=22\)
  • \((x−3)(x−5)=24\)
  • \(−2x(x−9)=x+21\)
  • \((x+1)(6x+1)=2x\)
  • \((x−2)(x+12)=15x\)
  • \((x+1)(x+2)=2(x+16)\)
  • \((x−9)(2x+3)=2(x−9)\)

3. \(4, 6\)

5. \(4, 10\)

9. \(−\frac{1}{5}, 2\)

11. \(\frac{7}{3}\)

13. \(−\frac{3}{2}, 1\)

15. \(−8, \frac{1}{2}\)

17. \(−6, 6\)

19. \(−\frac{9}{2}, \frac{9}{2}\)

21. \(−2, 2\)

23. \(−\frac{5}{4}, \frac{5}{4}\)

25. \(−3, 3\)

27. \(−2, 0\)

29. \(0, 2\)

31. \(−6, 4\)

33. \(\frac{1}{5}, 4\)

35. \(−1, 9\)

37. \(−\frac{1}{2}, −\frac{1}{3}\)

39. \(−6, 5\)

Exercise \(\PageIndex{8}\) Solve by Factoring

Clear the fractions by first multiplying both sides by the LCD and then solve.

  • \(115x^{2}+13x+25=0\)
  • \(114x^{2}−12x+37=0\)
  • \(32x^{2}−23=0\)
  • \(52x^{2}−110=0\)
  • \(314x^{2}−212=0\)
  • \(13x^{2}−15x=0\)
  • \(132x^{2}−12x+2=0\)
  • \(13x^{2}+56x−12=0\)
  • The sides of a square measure \(x + 3\) units. If the area is \(25\) square units, then find \(x\).
  • The height of a triangle is \(2\) units more than its base. If the area is \(40\) square units, then find the length of the base.
  • The sides of a right triangle have measures that are consecutive integers. Find the length of the hypotenuse. (Hint: The hypotenuse is the longest side. Apply the Pythagorean theorem.)
  • The profit in dollars generated by producing and selling \(x\) custom lamps is given by the function \(P(x)=−10x^{2}+800x−12000\). How many lamps must be sold and produced to break even? (Hint: We break even when the profit is zero.)

1. \(−3, −2\)

3. \(−\frac{2}{3}, \frac{2}{3}\)

5. \(±7\)

9. \(2\) units

11. \(5\) units

Exercise \(\PageIndex{9}\) Solve by Factoring

Assuming dry road conditions and average reaction times, the safe stopping distance, \(d\) in feet of an average car is given using the formula \(d=120v^{2}+v\), where \(v\) represents the speed of the car in miles per hour. For each problem below, given the stopping distance, determine the safe speed.

  • \(15\) feet
  • \(40\) feet
  • \(75\) feet
  • \(120\) feet

1. \(10\) miles per hour

3. \(30\) miles per hour

Exercise \(\PageIndex{10}\) Finding Equations with Given Solutions

Find a quadratic equation with integer coefficients, given the following solutions.

  • \(−3, 1\)
  • \(−5, 3\)
  • \(−10, −3\)
  • \(−7, −4\)
  • \(−1, 0\)
  • \(0, \frac{3}{5}\)
  • \(−2, 2\)
  • \(−\frac{1}{2}, \frac{1}{2}\)
  • \(−4, \frac{1}{3}\)
  • \(\frac{2}{3}, \frac{2}{5}\)
  • \(−\frac{1}{5}, −\frac{2}{3}\)
  • \(−\frac{3}{2}, \frac{3}{4}\)
  • \(3\), double root
  • \(−5\), double root

1. \(x^{2}+2x−3=0\)

3. \(x^{2}+13x+30=0\)

5. \(x^{2}+x=0\)

7. \(x^{2}−4=0\)

9. \(3x^{2}+11x−4=0\)

11. \(15x^{2}+13x+2=0\)

13. \(x^{2}−6x+9=0\)

Exercise \(\PageIndex{11}\) Solving Polynomial Equations

  • \(7x(x+5)(x−9)=0\)
  • \((x−1)(x−2)(x−3)=0\)
  • \(−2x(x−10)(x−1)=0\)
  • \(8x(x−4)^{2}=0\)
  • \(4(x+3)(x−2)(x+1)=0\)
  • \(−2(3x+1)(3x−1)(x−1)(x+1)=0\)
  • \(x^{3}−x^{2}−2x=0\)
  • \(2x^{3}+5x^{2}−3x=0\)
  • \(5x^{3}−15x^{2}+10x=0\)
  • \(−2x^{3}+2x^{2}+12x=0\)
  • \(3x^{3}−27x=0\)
  • \(−2x^{3}+8x=0\)
  • \(x^{3}+x^{2}−x−1=0\)
  • \(x^{3}+2x^{2}−16x−32=0\)
  • \(8x^{3}−4x^{2}−18x+9=0\)
  • \(12x^{3}=27x\)

1. \(−5, 0, 9 \)

3. \(0, 1, 10 \)

5. \(−3, −1, 2\)

7. \(−1, 0, 2\)

9. \(0, 1, 2\)

11. \(−3, 0, 3\)

13. \(−1, 1\)

15. \(−\frac{3}{2}, \frac{1}{2}, \frac{3}{2}\)

Exercise \(\PageIndex{12}\) Discussion Board Topics

  • Explain why \(2(x+5)(x−5)=0\) has two solutions and \(2x(x+5)(x−5)=0\) has three solutions.
  • Make up your own quadratic equation and post it and the solutions on the discussion board.
  • Explain, in your own words, how to solve a quadratic equation in standard form.

1. Answers may vary

3. Answers may vary

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Factoring Polynomials

When numbers are multiplied together, each of the numbers multiplied to get the product is called a factor . Sometimes it is desirable to write a polynomial as the product of certain of its factors. This operation is called factoring . Here we are interested in factoring polynomials with integral coefficients.

A polynomial is said to be factored completely if it is expressed as the product of polynomials with integral coefficients, and no one of the factors can still be written as the product of two polynomials with integral coefficients.

Following is a discussion of factoring some special polynomials.

Factors Common to All Terms

The greatest common factor (GCF) of a set of integers is defined as the greatest integer that divides each number of that set of integers.

The GCF can be obtained as follows:

1.  Factor the integers into their prime factors.

2.   Write the factors in the exponent form.

3.   Take the common bases each to its lowest exponent.

Find the GCF of 30 , 45 , 60 .

60 = 2^2*3*5

The common bases are 3 and 5 .

The least exponent of 3 is 1 and of 5 is 1 .

Hence the GCF = 3^1*5^1 = 15 .

The greatest common factor of a set of monomials can be found by taking the product of the GCF of the coefficients of the monomials and the common literal bases. each to its lowest exponent.

Find the GCF of 9x^(3)y^(2) , 12x^4y , -15x^5 .

9x^3y^2=3^2x^3y^2

12x^4y=2^2*3x^4y

  -15x^5=-3*5x^5

The common bases are 3 and x .

The least exponent of 3 is 1 and of x is 3 .

Hence the GCF = 3x^3 .

Find the GCF of 6a^4(x - y)^2 , 9a^3(x - y)^3 , 12a^2(x - y)^4 .

6a^4(x - y)^2=2*3a^4(x-y)^2

9a^3(x - y)^3=3^(2)a^(3)(x-y)^3

12a^2(x - y)^4=2^2*3a^2(x-y)^4

The common bases are 3 , a , and (x - y) .

The least exponent of 3 is 1 , of a is 2 , and of (x - y) is 2 .

Hence the GCF = 3a^2(x - y)^2 .

Note Since (1 - x)=-(x - 1) , the GCF of a(x - 1) , b(1 - x) is either (x - 1) or (1-x) .

When the terms of a polynomial have a common factor, the distributive law,

ab_1+ab_2+ab_3+...+ab_n=a(b_1+b_2+b_3+...b_n)

is used to factor the polynomial. One factor is the greatest common factor of all the terms of the polynomial. The other factor is the entire quotient, obtained by dividing each term of the polynomial by the common factor; that is,

ab_1+ab_2+ab_3+...+ab_n=a((ab_1)/a+(ab_2)/a+(ab_3)/a...+(ab_n)/a)

= a(b_1+b_2+b_3+...b_n)

Factor the expression 3a^2 - a .

The greatest common factor is a .

3a^2 - a = a((3a^2)/a-a/a)

            = a(3a-1)

Factor the polynomial 6x^(3)y^(2) + 12x^(2)y^(2) - 24xy^(2) .

The greatest common factor is 6xy^2 .

6x^(3)y^(2) + 12x^(2)y^(2) - 24xy^(2) = 6xy^2((6x^(3)y^(2))/(6xy^2)+(12x^(2)y^(2))/(6xy^2)-(24xy^2)/(6xy^2)

                                         = 6xy^2(x^2+2x-4)

Factor the polynomial 4x^2(2x - 1) - 8x(2x - 1)^2 .

The greatest common factor is 4x(2x - 1) .

4x^2(2x - 1) - 8x(2x - 1)^2 = 4x(2x - 1)[(4x^2(2x-1))/(4x(2x-1))-(8x(2x-1)^2)/(4x(2x-1))

                                             = 4x(2x-1)[x-2(2x-1)

                                             = 4x(2x-1)[x-4x+2

                                             = 4x(2x-1)(2-3x)

This is how our factorization calculator solves the problem above. You can see similar problems solved by clicking on 'Solve similar' button.

Factoring a Binomial

The methods of factoring polynomials will be presented according to the number of terms in the polynomial to be factored.

A monomial is already in factored form; thus the first type of polynomial to be considered for factoring is a binomial. Here we shall discuss factoring one type of binomials.

Squares and Square Roots

The squares of the numbers      3 ,        5^2 ,        a ,        x^2 ,        and        b^3

are, respectively,                      3^2 ,      5^4 ,       a^2 ,       x^4 ,         and      b^6

The 3 , 5^2 , a , x^2 , and b^3 are called the square roots of 3^2 , 5^4 , a^2 , x^4 , and b^6 , respectively.

Square root of a

Although the square of both (+ 3) and (- 3) is 9 , when we talk about the square root of 9 , we will mean the positive number 3 and not the negative number (-3) .

A number is said to be a perfect square if its square root is a rational number.

The square root of a specific number can be found by factoring the number into its prime factors, writing it in the exponent form, and then taking each base to one-half of its original exponent (when we square a number, we multiply its exponent by 2 ).

1.  root(64) = root(2^6) = 2^3 = 8

2.   root(144) = root(2^(4)*3^(2) = 2^2*3^1 = 12

When a is a literal number and n {is-in} N , we define root(a^(2n)) as (root(a))^(2n) = a^n . If the exponent is not divisible by 2 , the number is not a perfect square.

1.  root(a^4) = a^2

2.   root(x^(2)y^(6)) = xy^3

3.   root(4x^(2)y^(4)) = 2xy^2

The numbers 2 , 3 , 5 , 7 , 8 , 10 etc., are not perfect square numbers. This means there are no rational numbers whose squares are 2 , 3 , 5 etc.

The square roots of numbers that are not perfect squares are called irrational numbers .

Difference of Two Squares

The product of the two factors (a + b)(a - b) is a^2 - b^2 , the difference of two perfect square terms. The factors of the difference of two squares are the sum and difference of the respective square roots of the two squares.

Factor 9a^2-4

The square root of 9a^2 is 3a and of 4 is 2 . Hence 9a^2 - 4=(3a + 2)(3a - 2)

Note Remember to factor the polynomial completely.

Factor completely x^4 - 81y^4 .

x^4 - 81y^4 = (x^2+9y^2)(x^2-9y^2)

               = (x^2+9y^2)(x+3y)(x-3y)

Note Before checking if the binomial is a difference of two squares, check for a common factor. That is always the first operation to be performed.

Factor completely 6x^4 - 6 .

6x^4 - 6 = 6(x^4-1)

            = 6(x^2+1)(x^2-1)

            = 6(x^2+1)(x+1)(x-1)

Note (a + b)(a - b) = (a - b)(a + b)

Factoring a Trinomial

Factoring trinomials is divided into two cases:

1.   When the trinomial is of the form x^2 + bx + c , b , c ∈ I , b!=0 , c!=0 .

2.   When the trinomial is of the form ax^2 + bx + c , a!=1 , a , b , c ∈ I , b!=0 , c!=0 .

Trinomials of the Form , b, c ∈ I , and b ≠ 0, c ≠ 0

Consider the following products:

      (x+m)(x+n) = x^2+(m+n)x+mn

      (x-m)(x-n) = x^2 + (-m - n)x + mn

      (x+m)(x-n) = x^2+(m-n)x-mn

      (x-m)(x+n) = x^2 + (-m + n)x - mn

We note the following relations between the products and their factors:

1.  The first term in each factor is the square root of the square term in the trinomial.

2.   The product of the second terms of the factors is the third term in the trinomial.

3.  The sum of the second terms, signed numbers, is the coefficient of the middle term in the trinomial.

Note    To find the second terms in the factors, look for two signed numbers whose product is the third term in the trinomial and whose sum is the coefficient of the middle term in the trinomial.

Note    When the sign of the third term in the trinomial is plus, the two signed numbers have like signs and are the same as the sign of the middle term in the trinomial.

Note   When the sign of the third term in the trinomial is minus, the two signed numbers have different signs, and the larger one numerically has the sign of the middle term in the trinomial.

Factor x^2+8x+15

The first term of each factor is root(x^2) = x

Hence  x^2+8x+15 = ( x           )( x            )

Since the sign of the last term (+15) is plus, the two signed numbers in the factors have like signs.

Since the sign of the middle term (+8x) is plus, the two signed numbers are positive.

x^2+8x+15 = ( x +          )( x +           )

We look for two natural numbers whose product is 15 and whose sum is 8 . The two numbers are 3 and 5 .

Hence x^2+8x+15 = (x+3)(x+5)

Factor x^2-10x+24

Hence x^2-10x+24 = ( x           )( x            )

Since the sign of the last term (+24) is plus, the two signed numbers in the factors have like signs.

Since the sign of the middle term (-10x) is minus, the two signed numbers are negative.

x^2-10x+24 = ( x -          )( x -           )

We look for two natural numbers whose product is 24 and whose sum is 10 . The two numbers are 4 and 6 .

Hence x^2-10x+24 = (x-4)(x-6)

Factor x^2-5x-36

x^2-5x-36 = ( x           )( x            )

Since the sign of the last term (-36) is minus, the two numbers in the factors have different signs.

x^2-5x-36 = ( x +          )( x -          )

Since the sign of the middle term (-5x) is minus, the numerically larger number has the negative sign.

x^2-5x-36 = ( x + smaller number)( x - larger number)

We look for two natural numbers whose product is 36 and whose difference is 5 . The two numbers are 4 and 9 .

Hence x^2-5x-36 = (x + 4)(x - 9)

You can check below how our factorization calculator factorize the trinomial above. You can see similar problems solved by clicking on 'Solve similar' button.

Factor x^2+3x-28

x^2+3x-28 = ( x           )( x            )

Since the sign of the last term (-28) is minus, the two numbers in the factors have different signs.

x^2+3x-28 = ( x +          )( x -           )

Since the sign of the middle term (+3x) is plus, the numerically larger number has the plus sign.

x^2+3x-28 = ( x + larger number)( x - smaller number)

We look for two natural numbers whose product is 28 and whose difference is 3 . The two numbers are 4 and 7 .

Hence x^2+3x-28 = (x+7)(x-4)

Factor (x - y)^2 - 3(x - y) - 10 .

(x - y)^2 - 3(x - y) - 10 is of the form a^2 - 3a - 10 , whose factors are (a - 5)(a + 2) .

Hence (x - y)^2 - 3(x - y) - 10 = [(x - y) - 5][(x - y) + 2

                                              = (x-y-5)(x-y+2)

Note       When the third term of the trinomial is a large number and its factors are not obvious, write the number as the product of its prime factors; then make products

              of factors using combinations of the primes.

Note       (x + a)(x + b) = (x + b)(x + a)

Trinomials of the Form , a ≠ 1, a, b, c ∈ I , b ≠ 0, c ≠ 0

Consider the product

     (2x + 4)(x + 3) = 2x^2 + 10x + 12

The first factor on the left contains the common factor 2 :

     2x + 4 = 2(x + 2)

Also, the expanded product contains the common factor 2 :

     2x^2 + 10x +12 = 2(x^2 + 5x + 6)

In general, if a factor of a product contains a common factor, then the expanded product will also contain that common factor.

On the other hand, if no factor in a product, (x + 5)(3x - 2) , contains a common factor, then the expanded product, 3x^2 + 13x - 10 , will not have a common factor. Conversely, if the terms of a product do not have a common factor, then neither will any of its factors.

In order to learn how to factor a trinomial of the form ax^2 + bx + c , let us first look at how we multiply two factors together to get a product of this form.

We multiply (2x + 3)(4x - 5) as follows:

multiplying two binomials

Let us go over the same multiplication again, as shown in Figure 6.1.

how to multiply binomials

The sum of the products in the direction of the arrows,

cross multiplication

is the middle term of the trinomial.

The following example illustrates how to use the scissors in factoring a trinomial

ax^2+bx+c , a!=1 , a , b , c ∈ I .

Factor 6x^2-5x-6 .

Find all possible pairs of factors whose product is the first term of , the trinomial; each factor must contain the square root of the literal number. Write these factors at the left side of the scissors.

multiplying different terms

Find all possible pairs of factors whose product is the third term of the trinomial, disregarding the signs, and write them at the right side of the scissors.

Write all possible arrangements using the factors of the first term and the factors of the third term.

trying different combinations when factoring

The eight scissors just shown give all possible arrangements of the factors of the first term of the trinomial and the factors of the third term of the trinomial.

The terms on top of the scissors form the first factor of the product, and the terms on the bottom of the scissors form the second factor of the product.

Since there is no common factor in the trinomial. there should not be a common factor between the terms at the top of the scissors or a common factor between the terms at the bottom of the scissors. If there is a common factor between the terms at the top or the terms at the bottom in an arrangement. that arrangement cannot be the correct one. Arrangements (1) , (3) , (4) , (5) , (6) and (7) have common factors, and we eliminate them.

The candidates are now limited to two arrangements.

multiplying when factoring

The middle term of the trinomial. which is the sum of the products in the direction of the arrows. will indicate which arrangement is the correct one.

Since the first arrangement gives x and 36x for the middle term, which cannot give a sum of -5x , the first arrangement is not the correct one. The second arrangement gives 9x and 4x for the middle term, and by taking the 9x with a minus sign and 4x with a plus sign, we get -9x + 4x = -5x .

Hence the correct arrangement is

cross multiplying when factoring

The factors of the first term of the trinomial are always taken positive. Thus, in order to arrive at -9x , the 3 at the right side of the scissors has to be taken negative while the 2 has to be taken positive to arrive at +4x . The complete arrangement is

checking the factorization

Hence 6x^2 - 5x - 6 = (2x - 3)(3x + 2)

Note       When the trinomial has a common factor, factor it first before you attempt factoring by the scissors.

Note       There is no reason to write any arrangement with a common factor between the top terms or a common factor between the bottom terms.

Note       When the coefficient of the first term, or the third term of the trinomial is a large number, write the number as the product of its prime factors, and form products of factors using combinations of the primes.

See how our factorization calculator solve the trinomial above. You can see similar problems solved by clicking on 'Solve similar' button.

Factor 6x^2+19x+15 .

result of cross multiplication

Hence  6x^2+19x+15 = (2x+3)(3x+5)

Factor 36x^4-241x^2+100 .

cross multiplying and adding

Hence  36x^4-241x^2+100 = (4x^2-25)(9x^2-4)

                                          = (2x+5)(2x-5)(3x+2)(3x-2)

Math Topics

More solvers.

  • Add Fractions
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Solving Polynomials

"Solving" means finding the "roots" ...

... a "root" (or "zero") is where the function is equal to zero :

In between the roots the function is either entirely above, or entirely below, the x-axis

Example: −2 and 2 are the roots of the function x 2 − 4

Let's check:

  • when x = −2, then x 2 − 4 = (−2) 2 − 4 = 4 − 4 = 0
  • when x = 2, then x 2 − 4 = 2 2 − 4 = 4 − 4 = 0

How do we solve polynomials? That depends on the Degree !

The first step in solving a polynomial is to find its degree.

The Degree of a Polynomial with one variable is ...

... the largest exponent of that variable.

When we know the degree we can also give the polynomial a name:

How To Solve

So now we know the degree, how to solve?

  • Read how to solve Linear Polynomials (Degree 1) using simple algebra.
  • Read how to solve Quadratic Polynomials (Degree 2) with a little work,
  • It can be hard to solve Cubic (degree 3) and Quartic (degree 4) equations,
  • And beyond that it can be impossible to solve polynomials directly.

So what do we do with ones we can't solve? Try to solve them a piece at a time!

If we find one root, we can then reduce the polynomial by one degree (example later) and this may be enough to solve the whole polynomial.

Here are some main ways to find roots.

1. Basic Algebra

We may be able to solve using basic algebra:

Example: 2x+1

2x+1 is a linear polynomial:

The graph of y = 2x+1 is a straight line

It is linear so there is one root.

Use Algebra to solve:

A "root" is when y is zero: 2x+1 = 0

Subtract 1 from both sides: 2x = −1

Divide both sides by 2: x = −1/2

And that is the solution:

(You can also see this on the graph)

We can also solve Quadratic Polynomials using basic algebra (read that page for an explanation).

2. By experience, or simply guesswork.

It is always a good idea to see if we can do simple factoring:

Example: x 3 +2x 2 −x

This is cubic ... but wait ... we can factor out "x":

x 3 +2x 2 −x = x(x 2 +2x−1)

Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.

Example: x 3 −8

Again this is cubic ... but it is also the " difference of two cubes ":

x 3 −8 = x 3 −2 3

And so we can turn it into this:

x 3 −8 = (x−2)(x 2 +2x+4)

There is a root at x=2, because:

(2−2)(2 2 +2×2+4) = (0) (2 2 +2×2+4)

And we can then solve the quadratic x 2 +2x+4 and we are done

3. Graphically.

Graph the polynomial and see where it crosses the x-axis.

Graphing is a good way to find approximate answers, and we may also get lucky and discover an exact answer.

Caution: before you jump in and graph it, you should really know How Polynomials Behave , so you find all the possible answers!

This is useful to know: When a polynomial is factored like this:

f(x) = (x−a)(x−b)(x−c)...

Then a, b, c, etc are the roots !

So Linear Factors and Roots are related, know one and we can find the other.

(Read The Factor Theorem for more details.)

Example: f(x) = (x 3 +2x 2 )(x−3)

We see "(x−3)", and that means that 3 is a root (or "zero") of the function.

Well, let us put "3" in place of x:

f(x) = (3 3 +2·3 2 )(3−3)

f(3) = (3 3 +2·3 2 )( 0 )

Yes! f(3)=0, so 3 is a root.

How to Check

Found a root? Check it!

Simply put the root in place of "x": the polynomial should be equal to zero.

Example: 2x 3 −x 2 −7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2 . We can check easily, just put "2" in place of "x":

f(2) = 2(2) 3 −(2) 2 −7(2)+2 = 16−4−14+2 = 0

Yes! f(2)=0 , so we have found a root!

How about where it crosses near −1.8 :

f(−1.8) = 2(−1.8) 3 −(−1.8) 2 −7(−1.8)+2 = −11.664−3.24+12.6+2 = −0.304

No, it isn't equal to zero, so −1.8 will not be a root (but it may be close!)

But we did discover one root, and we can use that to simplify the polynomial, like this

Example (continued): 2x 3 −x 2 −7x+2

So, f(2)=0 is a root ... that means we also know a factor:

(x−2) must be a factor of 2x 3 −x 2 −7x+2

Next, divide 2x 3 −x 2 −7x+2 by (x−2) using Polynomial Long Division to find:

2x 3 −x 2 −7x+2 = (x−2)(2x 2 +3x−1)

So now we can solve 2x 2 +3x−1 as a Quadratic Equation and we will know all the roots.

That last example showed how useful it is to find just one root. Remember:

If we find one root, we can then reduce the polynomial by one degree and this may be enough to solve the whole polynomial.

How Far Left or Right

When trying to find roots, how far left and right of zero should we go?

There is a way to tell, and there are a few calculations to do, but it is all simple arithmetic. Read Bounds on Zeros for all the details.

Have We Got All The Roots?

There is an easy way to know how many roots there are. The Fundamental Theorem of Algebra says:

A polynomial of degree n ... ... has n roots (zeros) but we may need to use complex numbers

So: number of roots = the degree of polynomial .

Example: 2x 3 + 3x − 6

The degree is 3 (because the largest exponent is 3), and so:

There are 3 roots.

But Some Roots May Be Complex

Yes, indeed, some roots may be complex numbers (ie have an imaginary part), and so will not show up as a simple "crossing of the x-axis" on a graph.

But there is an interesting fact:

Complex Roots always come in pairs !

So we either get no complex roots, or 2 complex roots, or 4 , etc... Never an odd number.

Which means we automatically know this:

Positive or Negative Roots?

There is also a special way to tell how many of the roots are negative or positive called the Rule of Signs that you may like to read about.

Multiplicity of a Root

Sometimes a factor appears more than once. We call that Multiplicity :

  • Multiplicity is how often a certain root is part of the factoring.

Example: f(x) = (x−5) 3 (x+7)(x−1) 2

This could be written out in a more lengthy way like this:

f(x) = (x−5)(x−5)(x−5)(x+7)(x−1)(x−1)

(x−5) is used 3 times, so the root "5" has a multiplicity of 3 , likewise (x+7) appears once and (x−1) appears twice. So:

  • the root +5 has a multiplicity of 3
  • the root −7 has a multiplicity of 1 (a "simple" root)
  • the root +1 has a multiplicity of 2

Q: Why is this useful? A: It makes the graph behave in a special way!

When we see a factor like (x-r) n , "n" is the multiplicity, and

  • even multiplicity just touches the axis at "r" (and otherwise stays one side of the x-axis)
  • odd multiplicity crosses the axis at "r" (changes from one side of the x-axis to the other)

We can see it on this graph:

Solver Title

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  • Middle School Math Solutions – Polynomials Calculator, Factoring Quadratics Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Factoring is the process... Read More

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Unit 10: Polynomial expressions, equations, & functions

About this unit, intro to polynomials.

  • Polynomials intro (Opens a modal)
  • The parts of polynomial expressions (Opens a modal)
  • Evaluating polynomials (Opens a modal)
  • Simplifying polynomials (Opens a modal)
  • Polynomials intro 4 questions Practice

Adding & subtracting polynomials

  • Adding polynomials (Opens a modal)
  • Subtracting polynomials (Opens a modal)
  • Polynomial subtraction (Opens a modal)
  • Adding & subtracting multiple polynomials (Opens a modal)
  • Adding polynomials (old) (Opens a modal)
  • Adding and subtracting polynomials review (Opens a modal)
  • Add polynomials (intro) 4 questions Practice
  • Subtract polynomials (intro) 4 questions Practice
  • Add & subtract polynomials 4 questions Practice

Adding & subtracting polynomials: two variables

  • Adding polynomials: two variables (intro) (Opens a modal)
  • Subtracting polynomials: two variables (intro) (Opens a modal)
  • Subtracting polynomials: two variables (Opens a modal)
  • Finding an error in polynomial subtraction (Opens a modal)
  • Polynomials review (Opens a modal)
  • Adding and subtracting polynomials with two variables review (Opens a modal)
  • Add & subtract polynomials: two variables (intro) 4 questions Practice
  • Add & subtract polynomials: two variables 4 questions Practice
  • Add & subtract polynomials: find the error 4 questions Practice

Multiplying monomials

  • Multiplying monomials (Opens a modal)
  • Multiplying monomials to find area: two variables (Opens a modal)
  • Multiplying monomials to find area (Opens a modal)
  • Multiplying monomials challenge (Opens a modal)
  • Multiplying monomials review (Opens a modal)
  • Multiply monomials 4 questions Practice
  • Multiply monomials (advanced) 4 questions Practice

Multiplying monomials by polynomials

  • Multiplying monomials by polynomials: area model (Opens a modal)
  • Multiplying monomials by polynomials (Opens a modal)
  • Multiplying monomials by polynomials challenge (Opens a modal)
  • Multiplying monomials by polynomials review (Opens a modal)
  • Multiply monomials by polynomials: area model 4 questions Practice
  • Multiply monomials by polynomials 4 questions Practice
  • Multiply monomials by polynomials challenge 4 questions Practice

Multiplying binomials

  • Multiplying binomials: area model (Opens a modal)
  • Multiplying binomials intro (Opens a modal)
  • Multiplying binomials (Opens a modal)
  • Multiplying binomials review (Opens a modal)
  • Multiply binomials: area model 4 questions Practice
  • Warmup: multiplying binomials intro 4 questions Practice
  • Multiply binomials intro 4 questions Practice
  • Multiply binomials 4 questions Practice

Special products of binomials

  • Special products of the form (x+a)(x-a) (Opens a modal)
  • Squaring binomials of the form (x+a)² (Opens a modal)
  • Special products of the form (ax+b)(ax-b) (Opens a modal)
  • Squaring binomials of the form (ax+b)² (Opens a modal)
  • Special products of binomials: two variables (Opens a modal)
  • More examples of special products (Opens a modal)
  • Squaring a binomial (old) (Opens a modal)
  • Binomial special products review (Opens a modal)
  • Multiply difference of squares 4 questions Practice
  • Polynomial special products: perfect square 4 questions Practice

Multiplying binomials by polynomials

  • Multiplying binomials by polynomials (Opens a modal)
  • Multiplying binomials by polynomials: area model (Opens a modal)
  • Multiplying binomials by polynomials challenge (Opens a modal)
  • Multiplying binomials by polynomials review (Opens a modal)
  • Multiplying binomials by polynomials (old) (Opens a modal)
  • Multiplying binomials with radicals (old) (Opens a modal)
  • Multiply binomials by polynomials 4 questions Practice

Polynomials word problems

  • Polynomial multiplication word problem (Opens a modal)
  • Polynomial word problem: rectangle and circle area (Opens a modal)
  • Polynomial word problem: total value of bills (Opens a modal)
  • Polynomial word problem: area of a window (Opens a modal)

Introduction to factorization

  • Intro to factors & divisibility (Opens a modal)
  • Factors & divisibility 4 questions Practice

Factoring monomials

  • Which monomial factorization is correct? (Opens a modal)
  • Factoring monomials (Opens a modal)
  • Worked example: finding the missing monomial factor (Opens a modal)
  • Worked example: finding missing monomial side in area model (Opens a modal)
  • Greatest common factor of monomials (Opens a modal)
  • Factor monomials 4 questions Practice
  • Greatest common factor of monomials 4 questions Practice

Factoring polynomials by taking common factors

  • Factoring with the distributive property (Opens a modal)
  • Factoring polynomials by taking a common factor (Opens a modal)
  • Taking common factor from binomial (Opens a modal)
  • Taking common factor from trinomial (Opens a modal)
  • Taking common factor: area model (Opens a modal)
  • Factoring polynomials: common binomial factor (Opens a modal)
  • Factoring by common factor review (Opens a modal)
  • Factoring polynomials: common factor (old) (Opens a modal)
  • Factor polynomials: common factor 4 questions Practice

Evaluating expressions with unknown variables

  • Worked example: evaluating expressions using structure (Opens a modal)
  • Worked example: evaluating expressions using structure (more examples) (Opens a modal)
  • Evaluate expressions using structure 4 questions Practice

Factoring quadratics intro

  • Factoring quadratics as (x+a)(x+b) (Opens a modal)
  • Factoring quadratics: leading coefficient = 1 (Opens a modal)
  • Factoring quadratics as (x+a)(x+b) (example 2) (Opens a modal)
  • More examples of factoring quadratics as (x+a)(x+b) (Opens a modal)
  • Factoring simple quadratics review (Opens a modal)
  • Warmup: factoring quadratics intro 4 questions Practice
  • Factoring quadratics intro 4 questions Practice

Factoring quadratics by grouping

  • Intro to grouping (Opens a modal)
  • Factoring by grouping (Opens a modal)
  • Factoring quadratics by grouping (Opens a modal)
  • Factoring quadratics: leading coefficient ≠ 1 (Opens a modal)
  • Factor quadratics by grouping 4 questions Practice

Factoring polynomials with quadratic forms

  • Factoring quadratics: common factor + grouping (Opens a modal)
  • Factoring quadratics: negative common factor + grouping (Opens a modal)
  • Factoring two-variable quadratics (Opens a modal)
  • Factoring two-variable quadratics: rearranging (Opens a modal)
  • Factoring two-variable quadratics: grouping (Opens a modal)
  • Factoring quadratics with common factor (old) (Opens a modal)
  • Factor polynomials: quadratic methods 4 questions Practice
  • Factor polynomials: quadratic methods (challenge) 4 questions Practice

Factoring quadratics: Difference of squares

  • Difference of squares intro (Opens a modal)
  • Factoring quadratics: Difference of squares (Opens a modal)
  • Factoring difference of squares: leading coefficient ≠ 1 (Opens a modal)
  • Factoring difference of squares: analyzing factorization (Opens a modal)
  • Factoring difference of squares: missing values (Opens a modal)
  • Factoring difference of squares: shared factors (Opens a modal)
  • Difference of squares intro 4 questions Practice
  • Difference of squares 4 questions Practice

Factoring quadratics: Perfect squares

  • Perfect square factorization intro (Opens a modal)
  • Factoring quadratics: Perfect squares (Opens a modal)
  • Factoring perfect squares (Opens a modal)
  • Identifying perfect square form (Opens a modal)
  • Factoring higher-degree polynomials: Common factor (Opens a modal)
  • Factoring perfect squares: negative common factor (Opens a modal)
  • Factoring perfect squares: missing values (Opens a modal)
  • Factoring perfect squares: shared factors (Opens a modal)
  • Perfect squares intro 4 questions Practice
  • Perfect squares 4 questions Practice

Strategy in factoring quadratics

  • Strategy in factoring quadratics (part 1 of 2) (Opens a modal)
  • Strategy in factoring quadratics (part 2 of 2) (Opens a modal)
  • Factoring quadratics in any form (Opens a modal)

Factoring polynomials with special product forms

  • Factoring using the perfect square pattern (Opens a modal)
  • Factoring using the difference of squares pattern (Opens a modal)
  • Factoring difference of squares: two variables (example 2) (Opens a modal)
  • Factor polynomials using structure 4 questions Practice

Long division of polynomials

  • Intro to long division of polynomials (Opens a modal)
  • Dividing polynomials: long division (Opens a modal)

Synthetic division of polynomials

  • Intro to polynomial synthetic division (Opens a modal)
  • Dividing polynomials: synthetic division (Opens a modal)
  • Why synthetic division works (Opens a modal)

Practice dividing polynomials with remainders

  • Divide polynomials by x (with remainders) (Opens a modal)
  • Divide polynomials by monomials (with remainders) (Opens a modal)
  • Dividing polynomials with remainders (Opens a modal)
  • Divide polynomials by monomials (with remainders) 4 questions Practice
  • Divide polynomials by linear expressions 4 questions Practice
  • Divide polynomials with remainders 4 questions Practice

Polynomial Remainder Theorem

  • Intro to the Polynomial Remainder Theorem (Opens a modal)
  • Remainder theorem: finding remainder from equation (Opens a modal)
  • Remainder theorem: checking factors (Opens a modal)
  • Remainder theorem: finding coefficients (Opens a modal)
  • Proof of the Polynomial Remainder Theorem (Opens a modal)
  • Remainder theorem 4 questions Practice

Binomial theorem

  • Intro to the Binomial Theorem (Opens a modal)
  • Pascal's triangle and binomial expansion (Opens a modal)
  • Expanding binomials (Opens a modal)
  • Expanding binomials w/o Pascal's triangle (Opens a modal)
  • Expand binomials 4 questions Practice

Understanding the binomial theorem

  • Binomial expansion & combinatorics (Opens a modal)
  • Pascal's triangle & combinatorics (Opens a modal)
  • Binomial expansion & combinatorics (old) (Opens a modal)

Advanced polynomial factorization methods

  • Factoring higher-degree polynomials (Opens a modal)
  • Factoring sum of cubes (Opens a modal)
  • Factoring difference of cubes (Opens a modal)

Proving polynomial identities

  • Analyzing polynomial identities (Opens a modal)
  • Polynomial identities 4 questions Practice

Polynomial identities with complex numbers

  • Complex numbers & sum of squares factorization (Opens a modal)
  • Factoring sum of squares (Opens a modal)
  • Factor polynomials: complex numbers 4 questions Practice

Quadratic equations with complex numbers

  • Solving quadratic equations: complex roots (Opens a modal)
  • Solve quadratic equations: complex solutions 4 questions Practice

Fundamental Theorem of Algebra

  • The Fundamental theorem of Algebra (Opens a modal)
  • Quadratics & the Fundamental Theorem of Algebra (Opens a modal)
  • Number of possible real roots of a polynomial (Opens a modal)

Finding zeros of polynomials

  • Finding zeros of polynomials (1 of 2) (Opens a modal)
  • Finding zeros of polynomials (2 of 2) (Opens a modal)
  • Finding zeros of polynomials (example 2) (Opens a modal)
  • Zeros of polynomials (with factoring) 4 questions Practice

Zeros of polynomials and their graphs

  • Zeros of polynomials & their graphs (Opens a modal)
  • Positive & negative intervals of polynomials (Opens a modal)
  • Zeros of polynomials (factored form) 4 questions Practice
  • Positive & negative intervals of polynomials 4 questions Practice

End behavior of polynomial functions

  • Intro to end behavior of polynomials (Opens a modal)
  • End behavior of polynomials (Opens a modal)
  • End behavior of functions & their graphs (Opens a modal)
  • End behavior of polynomials 4 questions Practice

Graphs of polynomials

  • Graphs of polynomials (Opens a modal)
  • Graphs of polynomials: Challenge problems (Opens a modal)

Introduction to symmetry of functions

  • Function symmetry introduction (Opens a modal)
  • Even and odd functions: Graphs (Opens a modal)
  • Even/odd functions & numbers (Opens a modal)
  • Even and odd functions: Graphs and tables 4 questions Practice

Symmetry of polynomial functions

  • Even & odd polynomials (Opens a modal)
  • Symmetry of polynomials (Opens a modal)
  • Even & odd functions: Equations 4 questions Practice

IMAGES

  1. Problem solving involving polynomial functions

    steps in problem solving involving factors of polynomials

  2. Problem solving involving polynomial functions

    steps in problem solving involving factors of polynomials

  3. Factoring Polynomials (Flowchart with Examples)

    steps in problem solving involving factors of polynomials

  4. Grade 8 Week 2 Solving Problems Involving Factors of Polynomials

    steps in problem solving involving factors of polynomials

  5. Solving Word Problems Involving Factors of Polynomials

    steps in problem solving involving factors of polynomials

  6. Math 8 Module 2: Solving Problems Involving Factors of Polynomials

    steps in problem solving involving factors of polynomials

VIDEO

  1. Problems Involving Factoring Polynomials

  2. Solving Polynomial Equations By Factoring and Using Synthetic Division

  3. Word Problems Involving Factoring Polynomials

  4. A-Level Maths: B6-13 [Polynomials: Solving Problems and Extending the Factor Theorem]

  5. Solving Polynomial Equations By Factoring and Using Synthetic Division

  6. How To Factor Polynomials The Easy Way!

COMMENTS

  1. 4.4: Solve Polynomial Equations by Factoring

    Solution. Step 1: Express the equation in standard form, equal to zero. In this example, subtract 5x from and add 7 to both sides. 15x2 + 3x − 8 = 5x − 7 15x2 − 2x − 1 = 0. Step 2: Factor the expression. (3x − 1)(5x + 1) = 0. Step 3: Apply the zero-product property and set each variable factor equal to zero.

  2. How to Factor Polynomials (Step-by-Step)

    In this case, we have to factor the cubic polynomial 3y³ + 18y² + y + 6 using the same grouping method as the previous example. Step One: Split the cubic polynomial into groups of two binomials. Start by splitting the cubic polynomial into two groups (two separate binomials).

  3. 1.5: Factoring Polynomials

    Figure 1.5.1 1.5. 1. The area of the entire region can be found using the formula for the area of a rectangle. A = lw = 10x × 6x = 60x2 units2 A = l w = 10 x × 6 x = 60 x 2 u n i t s 2. The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region.

  4. 1.4: Solving Polynomial Equations

    Solve the linear equations. Check. Substitute each solution separately into the original equation. Zero of a Function: For any function \(f\), if \(f(x)=0\), then \(x\) is a zero of the function. How to use a problem solving strategy to solve word problems. Read the problem. Make sure all the words and ideas are understood. Identify what we are ...

  5. Solve Polynomial Equations by Factoring

    In this example, subtract 5x from and add 7 to both sides. 15x2 + 3x − 8 = 5x − 7 15x2 − 2x − 1 = 0. Step 2: Factor the expression. (3x − 1)(5x + 1) = 0. Step 3: Apply the zero-product property and set each variable factor equal to zero. 3x − 1 = 0 or 5x + 1 = 0. Step 4: Solve the resulting linear equations.

  6. Solving quadratic equations by factoring (article)

    This is how the solution of the equation 2 x 2 − 12 x + 18 = 0 goes: 2 x 2 − 12 x + 18 = 0 x 2 − 6 x + 9 = 0 Divide by 2. ( x − 3) 2 = 0 Factor. ↓ x − 3 = 0 x = 3. All terms originally had a common factor of 2 , so we divided all sides by 2 —the zero side remained zero—which made the factorization easier.

  7. 6.4 General Strategy for Factoring Polynomials

    2.1 Use a General Strategy to Solve Linear Equations; 2.2 Use a Problem Solving Strategy; 2.3 Solve a Formula for a Specific Variable; 2.4 Solve Mixture and Uniform Motion Applications; 2.5 Solve Linear Inequalities; 2.6 Solve Compound Inequalities; 2.7 Solve Absolute Value Inequalities

  8. Factoring polynomials by taking a common factor

    Factoring out the greatest common factor (GCF) To factor the GCF out of a polynomial, we do the following: Find the GCF of all the terms in the polynomial. Express each term as a product of the GCF and another factor. Use the distributive property to factor out the GCF. Let's factor the GCF out of 2 x 3 − 6 x 2 .

  9. Polynomial factorization: FAQ (article)

    The formula for finding the sum of an infinite geometric series is a / (1 - r), where a is the first term and r is the common ratio. If |r| < 1, then the sum of the series is finite and can be calculated using this formula. If |r| >= 1, then the series diverges and does not have a finite sum. For example, consider the geometric series 1 + 2 + 4 ...

  10. Algebra

    Section 1.5 : Factoring Polynomials. Of all the topics covered in this chapter factoring polynomials is probably the most important topic. There are many sections in later chapters where the first step will be to factor a polynomial. So, if you can't factor the polynomial then you won't be able to even start the problem let alone finish it.

  11. Algebra

    1.6 Multiplying Polynomials; 1.7 Factoring; 1.8 Simplifying Rational Expressions; 1.9 Graphing and Common Graphs; 1.10 Solving Equations, Part I; 1.11 Solving Equations, Part II; 1.12 Solving Systems of Equations; 1.13 Solving Inequalities; 1.14 Absolute Value Equations and Inequalities; 2. Trigonometry. 2.1 Trig Function Evaluation; 2.2 Graphs ...

  12. Factor a polynomial or an expression with Step-by-Step Math Problem Solver

    Steps 1 and 2 in this method are the same as in the previous method. Step 3 Rewrite the original problem by breaking the middle term into the two parts found in step 2. 8x - 5x = 3x, so we may write. Step 4 Factor this problem from step 3 by the grouping method studied in section 8-2.

  13. PDF Quarter 1 Module 2 Solving Problems Involving Factors of Polynomials

    Lesson 1- Solving Problems Involving Factors of Polynomials After going through this module, you are expected to: 1. recall the different techniques of factoring polynomials; 2. apply the concept of factoring in solving related problems; and 3. describe the importance of understanding factoring and its application to real-life;

  14. Polynomial factorization

    Algebra 2 12 units · 113 skills. Unit 1 Polynomial arithmetic. Unit 2 Complex numbers. Unit 3 Polynomial factorization. Unit 4 Polynomial division. Unit 5 Polynomial graphs. Unit 6 Rational exponents and radicals. Unit 7 Exponential models. Unit 8 Logarithms.

  15. How to Solve Polynomials: 13 Steps (with Pictures)

    A linear polynomial will have only one answer. If you need to solve a quadratic polynomial, write the equation in order of the highest degree to the lowest, then set the equation to equal zero. Rewrite the expression as a 4-term expression and factor the equation by grouping. Rewrite the polynomial as 2 binomials and solve each one. If you want ...

  16. 6.6: Solving Equations by Factoring

    The steps required to solve by factoring are outlined in the following example. Example \ (\PageIndex {4}\) Solve: \ (2x^ {2}+10x+20=−3x+5\). Solution: Step 1: Express the quadratic equation in standard form. For the zero-product property to apply, the quadratic expression must be equal to zero.

  17. Factoring Polynomials Step-by-Step Math Problem Solver

    The greatest common factor (GCF) of a set of integers is defined as the greatest integer that divides each number of that set of integers. The GCF can be obtained as follows: 1. Factor the integers into their prime factors. 2. Write the factors in the exponent form. 3. Take the common bases each to its lowest exponent.

  18. Solving Polynomials

    Use Algebra to solve: A "root" is when y is zero: 2x+1 = 0. Subtract 1 from both sides: 2x = −1. Divide both sides by 2: x = −1/2. And that is the solution: x = −1/2. (You can also see this on the graph) We can also solve Quadratic Polynomials using basic algebra (read that page for an explanation). 2.

  19. Factor Polynomials Calculator

    Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems

  20. Polynomial expressions, equations, & functions

    Test your understanding of Polynomial expressions, equations, & functions with these % (num)s questions. Start test. This topic covers: - Adding, subtracting, and multiplying polynomial expressions - Factoring polynomial expressions as the product of linear factors - Dividing polynomial expressions - Proving polynomials identities - Solving ...

  21. Solving Problems Involving Factors of Polynomials (English-Tagalog

    Hello my grade 8 students! I provided samples of how to solve problems involving factors of polynomials ^_^

  22. PDF Quarter 1 Module 2 Solving Problems Involving Factors of Polynomials

    equation that you are going to solve with accuracy and using variety of strategies. This module contains: Lesson 1- Solving Problems Involving Factors of Polynomials After going through this module, you are expected to: 1. recall the different methods of factoring polynomials; 2. apply the concept of factoring in solving related problems; and 3 ...

  23. Problem Solving Involving Factoring

    Problem Solving Involving Factoring - Download as a PDF or view online for free. Submit Search. ... Read the problems carefully and answer using your knowledge in factoring polynomials. Follow the steps suggested above for you to be guided. 1. The square of a number equals nine times that number. Find the number.