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9.7: Factoring Polynomials Completely

Created by: CK-12

Learning Objectives

  • Factor out a common binomial.
  • Factor by grouping.
  • Factor a quadratic trinomial where a \neq 1.
  • Solve real world problems using polynomial equations.

Introduction

We say that a polynomial is factored completely when we can’t factor it any more. Here are some suggestions that you should follow to make sure that you factor completely:

  • Factor all common monomials first.
  • Identify special products such as difference of squares or the square of a binomial. Factor according to their formulas.
  • If there are no special products, factor using the methods we learned in the previous sections.
  • Look at each factor and see if any of these can be factored further.

Example 1

Factor the following polynomials completely.

a) 6x^2-30x+24

b) 2x^2-8

c) x^3+6x^2+9x

Solution

a) Factor out the common monomial. In this case 6 can be divided from each term:

6(x^2-5x-6)

There are no special products. We factor x^2-5x+6 as a product of two binomials: (x \ )(x \ )

The two numbers that multiply to 6 and add to -5 are -2 and -3, so:

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

If we look at each factor we see that we can factor no more.

The answer is 6(x-2)(x-3).

b) Factor out common monomials: 2x^2-8=2(x^2-4)

We recognize x^2-4 as a difference of squares. We factor it as (x+2)(x-2).

If we look at each factor we see that we can factor no more.

The answer is 2(x+2)(x-2).

c) Factor out common monomials: x^3+6x^2+9x=x(x^2+6x+9)

We recognize x^2+6x+9 as a perfect square and factor it as (x+3)^2.

If we look at each factor we see that we can factor no more.

The answer is x(x+3)^2.

Example 2

Factor the following polynomials completely:

a) -2x^4+162

b) x^5-8x^3+16x

Solution

a) Factor out the common monomial. In this case, factor out -2 rather than 2. (It’s always easier to factor out the negative number so that the highest degree term is positive.)

-2x^4+162=-2(x^4-81)

We recognize expression in parenthesis as a difference of squares. We factor and get:

-2(x^2-9)(x^2+9)

If we look at each factor we see that the first parenthesis is a difference of squares. We factor and get:

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

If we look at each factor now we see that we can factor no more.

The answer is -2(x+3)(x-3)(x^2+9).

b) Factor out the common monomial: x^5-8x^3+14x=x(x^4-8x^2+16)

We recognize x^4-8x^2+16 as a perfect square and we factor it as x(x^2-4)^2.

We look at each term and recognize that the term in parentheses is a difference of squares.

We factor it and get ((x+2)(x-2))^2, which we can rewrite as (x+2)^2(x-2)^2.

If we look at each factor now we see that we can factor no more.

The final answer is x(x+2)^2(x-2)^2.

Factor out a Common Binomial

The first step in the factoring process is often factoring out the common monomials from a polynomial. But sometimes polynomials have common terms that are binomials. For example, consider the following expression:

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

Since the term (3x+2) appears in both terms of the polynomial, we can factor it out. We write that term in front of a set of parentheses containing the terms that are left over:

(3x+2)(x-5)

This expression is now completely factored.

Let’s look at some more examples.

Example 3

Factor out the common binomials.

a) 3x(x-1)+4(x-1)

b) x(4x+5)+(4x+5)

Solution

a) 3x(x-1)+4(x-1) has a common binomial of (x-1).

When we factor out the common binomial we get (x-1)(3x+4).

b) x(4x+5)+(4x+5) has a common binomial of (4x+5).

When we factor out the common binomial we get (4x+5)(x+1).

Factor by Grouping

Sometimes, we can factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factor by grouping.

The next example illustrates how this process works.

Example 4

Factor 2x+2y+ax+ay.

Solution

There is no factor common to all the terms. However, the first two terms have a common factor of 2 and the last two terms have a common factor of a. Factor 2 from the first two terms and factor a from the last two terms:

2x + 2y + ax + ay = 2(x + y) + a(x + y)

Now we notice that the binomial (x + y) is common to both terms. We factor the common binomial and get:

(x + y)(2 + a)

Example 5

Factor 3x^2+6x+4x+8.

Solution

We factor 3 from the first two terms and factor 4 from the last two terms:

3x(x+2)+4(x+2)

Now factor (x+2) from both terms: (x+2)(3x+4).

Now the polynomial is factored completely.

Factor Quadratic Trinomials Where a ≠ 1

Factoring by grouping is a very useful method for factoring quadratic trinomials of the form ax^2+bx+c, where a \neq 1.

A quadratic like this doesn’t factor as (x \pm m)(x \pm n), so it’s not as simple as looking for two numbers that multiply to c and add up to b. Instead, we also have to take into account the coefficient in the first term.

To factor a quadratic polynomial where a \neq 1, we follow these steps:

  1. We find the product ac.
  2. We look for two numbers that multiply to ac and add up to b.
  3. We rewrite the middle term using the two numbers we just found.
  4. We factor the expression by grouping.

Let’s apply this method to the following examples.

Example 6

Factor the following quadratic trinomials by grouping.

a) 3x^2+8x+4

b) 6x^2-11x+4

c) 5x^2-6x+1

Solution

Let’s follow the steps outlined above:

a) 3x^2+8x+4

Step 1: ac = 3 \cdot 4 = 12

Step 2: The number 12 can be written as a product of two numbers in any of these ways:

12 &= 1 \cdot 12 && \text{and} && 1 + 12 = 13\\12 &= 2 \cdot 6	&& \text{and} && 2 + 6 = 8 \qquad This \ is \ the \ correct \ choice.\\12 &= 3 \cdot 4 && \text{and} && 3 + 4 = 7

Step 3: Re-write the middle term: 8x = 2x + 6x, so the problem becomes:

3x^2+8x+4=3x^2+2x+6x+4

Step 4: Factor an x from the first two terms and a 2 from the last two terms:

x(3x+2)+2(3x+2) Now factor the common binomial (3x + 2):

(3x+2)(x+2) \qquad This \ is \ the \ answer.

To check if this is correct we multiply (3x+2)(x+2):

& \qquad \ \ 3x+2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;x+2\;}\\& \quad \quad \ \ 6x+4\\& \underline{3x^2+2x \;\;\;\;\;}\\& 3x^2+8x+4

The solution checks out.

b) 6x^2-11x+4

Step 1: ac = 6 \cdot 4 = 24

Step 2: The number 24 can be written as a product of two numbers in any of these ways:

24 &= 1 \cdot 24 && \text{and} && 1 + 24 = 25\\24 &= -1 \cdot (-24) && \text{and} && -1 + (-24) = -25\\24 &= 2 \cdot 12 && \text{and} && 2 + 12 = 14\\24 &= -2 \cdot (-12) && \text{and} && -2 + (-12) = -14\\24 &= 3 \cdot 8 && \text{and} && 3 + 8 = 11\\24 &= -3 \cdot (-8) && \text{and} && -3 + (-8) = -11 \qquad This \ is \ the \ correct \ choice.\\24 &= 4 \cdot 6	&& \text{and} && 4 + 6 = 10\\24 &= -4 \cdot (-6) && \text{and} && -4 + (-6) = -10

Step 3: Re-write the middle term: -11x = -3x - 8x, so the problem becomes:

6x^2-11x+4=6x^2-3x-8x+4

Step 4: Factor by grouping: factor a 3x from the first two terms and a -4 from the last two terms:

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

Now factor the common binomial (2x - 1):

(2x-1)(3x-4) \qquad This \ is \ the \ answer.

c) 5x^2-6x+1

Step 1: ac = 5 \cdot 1 = 5

Step 2: The number 5 can be written as a product of two numbers in any of these ways:

5 &= 1 \cdot 5 && \text{and} && 1 + 5 = 6\\5 &= -1 \cdot (-5) && \text{and} && -1 + (-5) = -6 \qquad This \ is \ the \ correct \ choice.

Step 3: Re-write the middle term: -6x = -x - 5x, so the problem becomes:

5x^2-6x+1=5x^2-x-5x+1

Step 4: Factor by grouping: factor an x from the first two terms and a - 1 from the last two terms:

x(5x-1)-1(5x-1)

Now factor the common binomial (5x - 1):

(5x-1)(x-1) \qquad This \ is \ the \ answer.

Solve Real-World Problems Using Polynomial Equations

Now that we know most of the factoring strategies for quadratic polynomials, we can apply these methods to solving real world problems.

Example 7

One leg of a right triangle is 3 feet longer than the other leg. The hypotenuse is 15 feet. Find the dimensions of the triangle.

Solution

Let x = the length of the short leg of the triangle; then the other leg will measure x + 3.

Use the Pythagorean Theorem: a^2+b^2=c^2, where a and b are the lengths of the legs and c is the length of the hypotenuse. When we substitute the values from the diagram, we get x^2+(x+3)^2=15^2.

In order to solve this equation, we need to get the polynomial in standard form. We must first distribute, collect like terms and rewrite in the form “polynomial = 0.”

x^2+x^2+6x+9& =225\\2x^2+6x+9& =225\\2x^2+6x-216 & =0

Factor out the common monomial: 2(x^2+3x-108)=0

To factor the trinomial inside the parentheses, we need two numbers that multiply to -108 and add to 3. It would take a long time to go through all the options, so let’s start by trying some of the bigger factors:

-108 &= -12 \cdot 9 && \text{and} && -12 + 9 = -3\\-108 &= 12 \cdot (-9) && \text{and} && 12 + (-9) = 3 \qquad This \ is \ the \ correct \ choice.

We factor the expression as 2(x-9)(x+12)=0.

Set each term equal to zero and solve:

& x-9=0 &&&& x+12=0\\& && \text{or}\\& \underline{\underline{x=9}} &&&& \underline{\underline{x=-12}}

It makes no sense to have a negative answer for the length of a side of the triangle, so the answer must be x = 9. That means the short leg is 9 feet and the long leg is 12 feet.

Check: 9^2+12^2=81+144=225=15^2, so the answer checks.

Example 8

The product of two positive numbers is 60. Find the two numbers if one numbers is 4 more than the other.

Solution

Let x = one of the numbers; then x + 4 is the other number.

The product of these two numbers is 60, so we can write the equation x(x+4)=60.

In order to solve we must write the polynomial in standard form. Distribute, collect like terms and rewrite:

x^2+4x &= 60\\x^2+4x-60 &= 0

Factor by finding two numbers that multiply to -60 and add to 4. List some numbers that multiply to -60:

-60 &= -4 \cdot 15 && \text{and} && -4 + 15 = 11\\-60 &= 4 \cdot (-15) && \text{and} && 4 + (-15) = -11\\-60 &= -5 \cdot 12 && \text{and} && -5 + 12 = 7\\-60 &= 5 \cdot (-12) && \text{and} && 5 + (-12) = -7\\-60 &= -6 \cdot 10 && \text{and} && -6 + 10 = 4 \qquad This \ is \ the \ correct \ choice.\\-60 & = 6 \cdot (-10) && \text{and} && 6 + (-10) = -4

The expression factors as (x+10)(x-6)=0.

Set each term equal to zero and solve:

& x+10=0 &&&& x-6=0\\& && \text{or}\\& \underline{\underline{x=-10}} &&&& \underline{\underline{x=6}}

Since we are looking for positive numbers, the answer must be x = 6. One number is 6, and the other number is 10.

Check: 6 \cdot 10 = 60, so the answer checks.

Example 9

A rectangle has sides of length x + 5 and x - 3. What is x if the area of the rectangle is 48?

Solution

Make a sketch of this situation:

Using the formula Area = length \times width, we have (x+5)(x-3)=48.

In order to solve, we must write the polynomial in standard form. Distribute, collect like terms and rewrite:

x^2+2x-15& =48\\x^2+2x-63& =0

Factor by finding two numbers that multiply to -63 and add to 2. List some numbers that multiply to -63:

-63 &= -7 \cdot 9 && \text{and} && -7 + 9 = 2 \qquad This \ is \ the \ correct \ choice.\\-63 &= 7 \cdot (-9) && \text{and} && 7 + (-9) = -2

The expression factors as (x+9)(x-7)=0.

Set each term equal to zero and solve:

& x+9=0 &&&& x-7=0\\& && \text{or}\\& \underline{\underline{x=-9}} &&&& \underline{\underline{x=7}}

Since we are looking for positive numbers the answer must be x = 7. So the width is x - 3 = 4 and the length is x + 5 = 12.

Check: 4 \cdot 12 = 48, so the answer checks.

Resources

The WTAMU Virtual Math Lab has a detailed page on factoring polynomials here: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm. This page contains many videos showing example problems being solved.

Review Questions

Factor completely.

  1. 2x^2+16x+30
  2. 5x^2-70x+245
  3. -x^3+17x^2-70x
  4. 2x^4-512
  5. 25x^4-20x^3+4x^2
  6. 12x^3+12x^2+3x

Factor by grouping.

  1. 6x^2-9x+10x-15
  2. 5x^2-35x+x-7
  3. 9x^2-9x-x+1
  4. 4x^2+32x-5x-40
  5. 2a^2-6ab+3ab-9b^2
  6. 5x^2+15x-2xy-6y

Factor the following quadratic trinomials by grouping.

  1. 4x^2+25x-21
  2. 6x^2+7x+1
  3. 4x^2+8x-5
  4. 3x^2+16x+21
  5. 6x^2-2x-4
  6. 8x^2-14x-15

Solve the following application problems:

  1. One leg of a right triangle is 7 feet longer than the other leg. The hypotenuse is 13. Find the dimensions of the right triangle.
  2. A rectangle has sides of x + 2 and x - 1. What value of x gives an area of 108?
  3. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other.
  4. A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is wide?
  5. Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the numbers?
  6. Two positive numbers have a sum of 8, and their product is equal to the smaller number plus 10. What are the numbers?
  7. Framing Warehouse offers a picture framing service. The cost for framing a picture is made up of two parts: glass costs $1 per square foot and the frame costs $2 per foot. If the frame has to be a square, what size picture can you get framed for $20?

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