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# 9.7: Factoring Polynomials Completely

Difficulty Level: At Grade 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?

Feb 23, 2012

Apr 02, 2015