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# 9.6: Factoring Special Products

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Factor the difference of two squares.
• Factor perfect square trinomials.
• Solve quadratic polynomial equation by factoring.

## Introduction

When you learned how to multiply binomials we talked about two special products.

$\text{The Sum and Difference Formula} && (a + b) (a - b ) = a^2 - b^2 \\\text{The Square of a Binomial Sormula} && (a + b)^2 = a^2 + 2ab + b^2 \\&& (a - b)^2 = a^2 - 2ab + b^2$

In this section we will learn how to recognize and factor these special products.

## Factor the Difference of Two Squares

We use the sum and difference formula to factor a difference of two squares. A difference of two squares can be a quadratic polynomial in this form.

$a^2 - b^2$

Both terms in the polynomial are perfect squares. In a case like this, the polynomial factors into the sum and difference of the square root of each term.

$a^2 - b^2 = (a + b) (a - b)$

In these problems, the key is figuring out what the $a$ and $b$ terms are. Let’s do some examples of this type.

Example 1

Factor the difference of squares.

a) $x^2 - 9$

b) $x^2 - 100$

c) $x^2 - 1$

Solution

a) Rewrite as $x^2 - 9$ as $x^2 - 3^2$. Now it is obvious that it is a difference of squares.

$\text{The difference of squares formula is} && a^2 - b^2 = (a + b)(a - b) \\\text{Letâ€™s see how our problem matches with the formula} && x^2 - 3^2 = (x + 3)(x - 3)$

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

We can check to see if this is correct by multiplying $(x + 3) (x - 3)$.

$& \quad \quad \quad \ x \ \ + \ 3\\& \ \ \underline{\;\;\;\;\;\;\;\;\;\;x \ \ - \ 3}\\& \quad \quad -3x \ - \ 9\\& \ \ \underline{x^2 \ + 3x\;\;\;\;\;\;\;\;\;}\\& \ \ x^2 \ + 0x \ - \ 9$

We could factor this polynomial without recognizing that it is a difference of squares. With the methods we learned in the last section we know that a quadratic polynomial factors into the product of two binomials.

$(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})$

We need to find two numbers that multiply to -9 and add to 0, since the middle term is missing.

We can write -9 as the following products

$& -9 = -1 \cdot 9 && \text{and} && -1 + 9 \ = 8 \\& -9 = 1 \cdot (-9) && \text{and} && 1 + (-9) = -8 \\& -9 = 3 \cdot (-3) && \text{and} && 3 + (-3) = 0 \qquad \leftarrow \qquad \text{This is the correct choice}$

We can factor $x^2 - 9$ as $(x + 3) (x - 3)$, which is the same answer as before.

You can always factor using methods for factoring trinomials, but it is faster if you can recognize special products such as the difference of squares.

b) Rewrite $x^2 - 100$ as $x^2 - 10^2$. This factors as $(x + 10) (x - 10)$.

c) Rewrite $x^2 - 1$ as $x^2 - 1^2$. This factors as $(x + 1) (x - 1)$.

Example 2

Factor the difference of squares.

a) $16x^2 - 25$

b) $4x^2 - 81$

c) $49x^2 - 64$

Solution

a) Rewrite $16x^2 - 25$ as $(4x)^2 - 5^2$. This factors as $(4x + 5) (4x - 5)$.

b) Rewrite $4x^2 - 81$ as $(2x)^2 - 9^2$. This factors as $(2x + 9) (2x - 9)$.

c) Rewrite $49x^2 - 64$ as $(7x)^2 - 8^2$. This factors as $(7x + 8) (7x - 8)$.

Example 3

Factor the difference of squares:

a) $x^2 - y^2$

b) $9x^2 - 4y^2$

c) $x^2y^2 - 1$

Solution

a) $x^2 - y^2$ factors as $(x + y) (x- y)$.

b) Rewrite $9x^2 - 4y^2$ as $(3x)^2 - (2y)^2$. This factors as $(3x + 2y) (3x - 2y)$.

c) Rewrite as $x^2y^2 - 1$ as $(xy)^2 - 1^2$. This factors as $(xy + 1) (xy - 1)$.

Example 4

Factor the difference of squares.

a) $x^4 - 25$

b) $16x^4 - y^2$

c) $x^2y^8 - 64z^2$

Solution

a) Rewrite $x^4 - 25$ as $(x^2)^2 - 5^2$. This factors as $(x^2 + 5) (x^2 - 5)$.

b) Rewrite $16x^4 - y^2$ as $(4x^2)^2 - y^2$. This factors as $(x^2 + 5) (x^2 - 5)$.

c) Rewrite $x^2y^8 - 64z^2$ as $(xy^2)^2 - (8z)$. This factors as $(xy^2 + 8z) (xy^2 - 8z)$.

## Factor Perfect Square Trinomials

We use the Square of a Binomial Formula to factor perfect square trinomials. A perfect square trinomial has the following form.

$a^2 + 2ab + b^2 \quad \quad \text{or} \quad \quad a^2 - 2ab + b^2$

In these special kinds of trinomials, the first and last terms are perfect squares and the middle term is twice the product of the square roots of the first and last terms. In a case like this, the polynomial factors into perfect squares.

$a^2 + 2ab + b^2 & = (a + b)^2 \\a^2 - 2ab + b^2 & = (a - b)^2$

In these problems, the key is figuring out what the a and b terms are. Let’s do some examples of this type.

Example 5

Factor the following perfect square trinomials.

a) $x^2 + 8x + 16$

b) $x^2 - 4x + 4$

c) $x^2 + 14x + 49$

Solution

a) $x^2 + 8x + 16$

The first step is to recognize that this expression is actually perfect square trinomials.

1. Check that the first term and the last term are perfect squares. They are indeed because we can re-write:

$x^2 + 8x + 16 \quad \quad \text{as} \quad \quad x^2 + 8x + 4^2.$

2. Check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them.

$x^2 + 8x + 16 \quad \quad \text{as} \quad \quad x^2 + 2 \cdot 4 \cdot x + 4^2$

This means we can factor $x^2 + 8x + 16$ as $(x + 4)^2$.

We can check to see if this is correct by multiplying $(x + 4) (x + 4).$

$& \quad \quad \quad \ x \ \ + \ \ 4\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;x \ \ - \ \ 4}\\& \quad \quad \quad 4x \ + \ 16\\& \underline{x^2 \ + \ 4x\;\;\;\;\;\;\;\;\;\;\;}\\& x^2 \ + \ 8x \ + \ 16$

We could factor this trinomial without recognizing it as a perfect square. With the methods we learned in the last section we know that a trinomial factors as a product of the two binomials in parentheses.

$(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})$

We need to find two numbers that multiply to 16 and add to 8. We can write 16 as the following products.

$&16 = 1 \cdot 16 && \text{and} && 1 + 16 = 17 \\&16 = 2 \cdot 8 && \text{and} && 2 + 8 = 10 \\&16 = 4 \cdot 4 && \text{and} && 4 + 4 = 8 \qquad \leftarrow \qquad \text{This is the correct choice.}$

We can factor $x^2 + 8x + 16$ as $(x + 4) (x + 4)$ which is the same as $(x + 4)^2$.

You can always factor by the methods you have learned for factoring trinomials but it is faster if you can recognize special products.

b) Rewrite $x^2 - 4x + 4$ as $x^2 + 2 \cdot (-2) \cdot x + (-2)^2$.

We notice that this is a perfect square trinomial and we can factor it as: $(x - 2)^2$.

c) Rewrite $x^2 + 14x + 49$ as $x^2 + 2 \cdot 7 \cdot x + 7^2.$

We notice that this is a perfect square trinomial as we can factor it as: $(x + 7)^2$.

Example 6

Factor the following perfect square trinomials.

a) $4x^2 + 20x + 25$

b) $9x^2 - 24x + 16$

c) $x + 2xy + y^2$

Solution

a) Rewrite $4x^2 + 20x + 25$ as $(2x)^2 + 2.5 \cdot (2x) + 5^2$

We notice that this is a perfect square trinomial and we can factor it as $(2x + 5)^2$.

b) Rewrite $9x^2 - 24x + 16$ as $(3x)^2 + 2 \cdot (-4) \cdot (3x) + (-4)^2.$

We notice that this is a perfect square trinomial as we can factor it as $(3x - 4)^2$.

We can check to see if this is correct by multiplying $(3x - 4)^2 = (3x - 4) (3x - 4)$.

$& \quad \quad \qquad 3x \ - \ 4\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3x \ - \ 4}\\& \quad \ \ - \ 12x \ + \ 16\\& \underline{9x^2 \ - \ 12x\;\;\;\;\;\;\;\;\;\;\;}\\& 9x^2 \ - \ 24x \ + \ 16$

c) $x + 2xy + y^2$

We notice that this is a perfect square trinomial as we can factor it as $(x + y)^2$.

## Solve Quadratic Polynomial Equations by Factoring

With the methods we learned in the last two sections, we can factor many kinds of quadratic polynomials. This is very helpful when we want to solve polynomial equations such as

$ax^2 + bx + c = 0$

Remember that to solve polynomials in expanded form we use the following steps:

Step 1

If necessary, rewrite the equation in standard form so that

Polynomial expression = 0.

Step 2

Factor the polynomial completely.

Step 3

Use the Zero-Product rule to set each factor equal to zero.

Step 4

Solve each equation from Step 3.

Step 5

We will do a few examples that show how to solve quadratic polynomials using the factoring methods we just learned.

Example 7

Solve the following polynomial equations.

a) $x^2 + 7x + 6 = 0$

b) $x^2 - 8x = -12$

c) $x^2 = 2x + 15$

Solution

a) $x^2 + 7x + 6 = 0$

Rewrite This is not necessary since the equation is in the correct form already.

Factor We can write 6 as a product of the following numbers.

$&6 = 1 \cdot 6 && \text{and} && 1 + 6 = 7 \qquad \leftarrow \qquad \text{This is the correct choice.}\\&6 = 2 \cdot 3 && \text{and} && 2 + 3 = 5$

$x^2 + 7x + 6 = 0$ factors as $(x + 1) (x + 6) = 0$

Set each factor equal to zero

$x + 1 = 0 \quad \quad \text{or} \quad \quad x + 6 = 0$

Solve

$x = -1 \quad \quad \text{or} \quad \quad x = -6$

Check Substitute each solution back into the original equation.

$x = -1 && (-1)^2 + 7(-1) + 6 = 1 - 7 + 6 = 0 && \ \text{Checks out.}\\x = -6 && (-6)^2 + 7(-6) + 6 = 36 - 42 + 6 = 0 && \ \text{Checks out.}$

b) $x^2 - 8x = -12$

Rewrite $x^2 - 8x = -12$ is rewritten as $x^2 - 8x + 12 = 0.$

Factor We can write 12 as a product of the following numbers.

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

$x^2 -8x + 12 = 0$ factors as $(x - 2) (x - 6) = 0$

Set each factor equal to zero.

$x - 2 = 0 \quad \quad \text{or} \quad \quad x - 6 = 0$

Solve.

$x = 2 \quad \quad \text{or} \quad \quad x = 6$

Check Substitute each solution back into the original equation.

$x = 2 && (2)^2 - 8(2) = 4 -16 = -12 && \ \text{Checks out.}\\x = 2 &&(6)^2 - 8(6) = 36 - 48 = -12 && \ \text{Checks out.}$

c) $x^2 = 2x + 15$

Rewrite $x^2 = 2x + 15$ is re-written as $x^2 - 2x - 15 = 0$.

Factor We can write -15 as a product of the following numbers.

$& -15 = 1 \cdot (-15) && \text{and} && 1 + (-15) = -14 \\& -15 = -1 \cdot (15) && \text{and} && -1 + (15) = 14 \\& -15 = -3 \cdot 5 && \text{and} && -3 + 5 = 2 \\& -15 = 3 \cdot (-5) && \text{and} && 3 + (-5) = -2 \qquad \leftarrow \qquad \text{This is the correct choice.}$

$x^2 - 2x - 15 = 0$ factors as $(x + 3) (x - 5) = 0$.

Set each factor equal to zero

$x + 3 = 0 \quad \quad \text{or} \quad \quad x - 5 = 0$

Solve

$x = -3 \quad \quad \text{or} \quad \quad x = 5$

Check Substitute each solution back into the original equation.

$x & = -3 && (-3)^2 = 2(-3) + 15 \Rightarrow 9 = 0 && \ \text{Checks out.}\\x & = 5 && (5)^2 = 2(5) + 15 \Rightarrow 25 = 25 && \ \text{Checks out}.$

Example 8

Solve the following polynomial equations.

a) $x^2 - 12x + 36 = 0$

b) $x ^2 - 81 = 0$

c) $x ^2 + 20x + 100 = 0$

Solution

a) $x^2 - 12x + 36 = 0$

Rewrite This is not necessary since the equation is in the correct form already.

Factor: Re-write $x^2 - 12x + 36$ as $x^2 - 2 \cdot (-6)x + (-6)^2.$

We recognize this as a difference of squares. This factors as $(x - 6)^2 = 0$ or $(x - 6) (x - 6) = 0$.

Set each factor equal to zero

$x - 6 = 0 \quad \quad \text{or} \quad \quad x - 6 = 0$

Solve

$x = 6 \quad \quad \text{or} \quad \quad x = 6$

Notice that for a perfect square the two solutions are the same. This is called a double root.

Check Substitute each solution back into the original equation.

$x = 6 && 6^2 - 12(6) + 36 = 36 - 72 + 36 + 0 && \ \text{Checks out.}$

b) $x^2 - 81 = 0$

Rewrite This is not necessary since the equation is in the correct form already

Factor Rewrite $x^2 - 81 = 0$ as $x^2 - 9^2 = 0.$

We recognize this as a difference of squares. This factors as $(x- 9) (x + 9) = 0$.

Set each factor equal to zero.

$x - 9 = 0 \quad \quad \text{or} \quad \quad x + 9 = 0$

Solve:

$x = 9 \quad \quad \text{or} \quad \quad x = -9$

Check: Substitute each solution back into the original equation.

$x & = 9 && 9^2 - 81 = 81 - 81 = 0 && \ \text{Checks out.}\\x & = -9 && (-9)^2 - 81 = 81 - 81 = 0 && \ \text{Checks out.}$

c) $x^2 + 20x + 100 = 0$

Rewrite This is not necessary since the equation is in the correct form already.

Factor Rewrite $x^2 + 20x + 100 = 0$ as $x^2 + 2 \cdot 10 \cdot x + 10^2$

We recognize this as a perfect square. This factors as: $(x + 10)^2 = 0$ or $(x + 10) (x +10) = 0.$

Set each factor equal to zero.

$x + 10 = 0 \quad \quad \text{or} \quad \quad x + 10 = 0$

Solve.

$x = -10 \quad \quad \text{or} \quad \quad x = -10 \quad \quad \text{This is a double root.}$

Check Substitute each solution back into the original equation.

$x = 10 && (-10)^2 + 20(-10) + 100 = 100 - 200 + 100 = 0 && \ \text{Checks out.}$

## Review Questions

Factor the following perfect square trinomials.

1. $x^2 + 8x + 16$
2. $x^2 - 18x + 81$
3. $-x^2 + 24x -144$
4. $x^2 + 14x + 49$
5. $4x^2 - 4x + 1$
6. $25x^2 + 60x + 36$
7. $4x^2 - 12xy + 9y^2$
8. $x^4 + 22x^2 + 121$

Factor the following difference of squares.

1. $x^2 - 4$
2. $x^2 - 36$
3. $-x^2 + 100$
4. $x^2 - 400$
5. $9x^2 - 4$
6. $25x^2 - 49$
7. $-36x^2 + 25$
8. $16x^2 - 81y^2$

Solve the following quadratic equation using factoring.

1. $x^2 - 11x + 30 = 0$
2. $x^2 + 4x = 21$
3. $x^2 + 49 = 14x$
4. $x^2 - 64 = 0$
5. $x^2 - 24x + 144 = 0$
6. $4x^2 - 25 = 0$
7. $x^2 + 26x = -169$
8. $-x^2 - 16x - 60 = 0$

1. $(x + 4)^2$
2. $(x - 9)^2$
3. $- (x - 12)^2$
4. $(x + 7)^2$
5. $(2x - 1)^2$
6. $(5x + 6)^2$
7. $(2x - 3y)^2$
8. $(x^2 + 11)^2$
9. $(x + 2) (x - 2)$
10. $(x + 6) (x - 6)$
11. $- (x + 10) (x - 10)$
12. $(x + 20) (x - 20)$
13. $(3x + 2) (3x - 2)$
14. $(5x + 7) (5x - 7)$
15. $- (6x + 5) (6x - 5)$
16. $(4x + 9y) (4x - 9y)$
17. $x = 5, x = 6$
18. $x = -7, x = 3$
19. $x = 7$
20. $x = -8, x = 8$
21. $x = 12$
22. $x = \frac{5}{2}, x = - \frac{5}{2}$
23. $x = -13$
24. $x = -10, x = -6$

Feb 22, 2012

Aug 22, 2014