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# 9.6: Factoring Special Products

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## 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.

$\text{The sum and difference formula:} \quad (a + b)(a - b) & = a^2 - b^2\\\text{The square of a binomial formulas:} \qquad \quad \ \ (a + b)^2 & = a^2 + 2ab + b^2\\ (a - b)^2 & = a^2 - 2ab + b^2$ In this section we’ll 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 is any quadratic polynomial in the form $a^2 - b^2$, where $a$ and $b$ can be variables, constants, or just about anything else. The factors of $a^2 - b^2$ are always $(a + b)(a - b)$; the key is figuring out what the $a$ and $b$ terms are.

Example 1

Factor the difference of squares:

a) $x^2 - 9$

b) $x^2 - 100$

c) $x^2 - 1$

Solution

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

$\text{The difference of squares formula is:} && a^2 - b^2 & = (a + b)(a - b)\\\text{Letâ€™s see how our problem matches with the formula:} && x^2 - 3^2 & = (x + 3)(x - 3)\\\text{The answer is:} && x^2 - 9 & = (x + 3)(x - 3)$

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

$& \quad \quad \ \ x + 3\\& \underline{\;\;\;\;\;\;\;\;\;x - 3}\\& \quad -3x - 9\\& \underline{x^2 + 3x\;\;\;\;\;\;}\\& x^2 + 0x - 9$ The answer checks out.

Note: We could factor this polynomial without recognizing it as 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:

$(x\;\;\;\;)(x\;\;\;\;)$

We need to find two numbers that multiply to -9 and add to 0 (since there is no $x-$term, that’s the same as if the $x-$term had a coefficient of 0). We can write -9 as the following products:

$& -9 = -1 \cdot 9 && \text{and} && -1 + 9 = 8\\& -9 = 1 \cdot (-9) && \text{and} && 1 + (-9) = -8\\& -9 = 3 \cdot (-3) && \text{and} && 3 + (-3) = 0 \qquad These \ are \ the \ correct \ numbers.$

We can factor $x^2 - 9$ as $(x + 3)(x - 3)$, which is the same answer as before. You can always factor using the methods you learned in the previous section, but recognizing special products helps you factor them faster.

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) $x^2 y^2 - 1$

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 $x^2 y^2 - 1$ 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) $x^2 y^8 - 64z^2$

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 $(4x^2 + y)(4x^2 - y)$.

c) Rewrite $x^2 y^4 - 64z^2$ as $(xy^2)^2 - (8z)^2$. 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 form $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$.

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:

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

Once again, the key is figuring out what the $a$ and $b$ terms are.

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) The first step is to recognize that this expression is a perfect square trinomial.

First, we can see that the first term and the last term are perfect squares. We can rewrite $x^2 + 8x + 16$ as $x^2 + 8x + 4^2$.

Next, we 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 $x^2 + 8x + 16$ as $x^2 + 2 \cdot 4 \cdot x + 4^2$.

This means we can factor $x^2 + 8x + 16$ as $(x + 4)^2$. We can check to see if this is correct by multiplying $(x + 4)^2 = (x + 4)(x + 4)$ :

$& \quad \quad \quad x + 4\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;x + 4}\\& \quad \quad \ 4x + 16\\& \underline{x^2 + 4x\;\;\;\;\;\;\;\;}\\& x^2 + 8x + 16$ The answer checks out.

Note: We could factor this trinomial without recognizing it as a perfect square. We know that a trinomial factors as a product of two binomials:

$(x\;\;\;\;)(x\;\;\;\;)$

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

$& 16 = 1 \cdot 16 && \text{and} && 1 + 16 = 17\\& 16 = 2 \cdot 8 && \text{and} && 2 + 8 = 10\\& 16 = 4 \cdot 4 && \text{and} && 4 + 4 = 8 \qquad These \ are \ the \ correct \ numbers$

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

Once again, you can factor perfect square trinomials the normal way, but recognizing them as perfect squares gives you a useful shortcut.

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, so 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, so 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) $x^2 + 2xy + y^2$

Solution

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

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 and we can factor it as $(3x - 4)^2$.

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

$& \quad \quad \quad 3x - 4\\& \underline{\;\;\;\;\;\;\;\;\;\;\;3x - 4\;\;}\\& \quad \ -12x + 16\\& \underline{9x^2 - 12x\;\;\;\;\;\;\;\;}\\& 9x^2 - 24x + 16$ The answer checks out.

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

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

For more examples of factoring perfect square trinomials, watch the videos at http://www.onlinemathlearning.com/perfect-square-trinomial.html.

## Solve Quadratic Polynomial Equations by Factoring

With the methods we’ve learned in the last two sections, we can factor many kinds of quadratic polynomials. This is very helpful when we want to solve them. Remember the process we learned earlier:

1. If necessary, rewrite the equation in standard form so that the right-hand side equals zero.
2. Factor the polynomial completely.
3. Use the zero-product rule to set each factor equal to zero.
4. Solve each equation from step 3.
5. Check your answers by substituting your solutions into the original equation

We can use this process 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) Rewrite: We can skip this since the equation is in the correct form already.

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

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

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

Set each factor equal to zero:

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

Solve:

$\underline{\underline{x = -1}} && \text{or} && \underline{\underline{x = -6}}$

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) 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:

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

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

Set each factor equal to zero:

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

Solve:

$\underline{\underline{x = 2}} && \text{or} && \underline{\underline{x = 6}}$

Check: Substitute each solution back into the original equation.

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

c) Rewrite: $x^2 = 2x + 15$ is rewritten as $x^2 - 2x - 15 = 0$

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

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

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

Set each factor equal to zero:

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

Solve:

$\underline{\underline{x = -3}} && \text{or} && \underline{\underline{x = 5}}$

Check: Substitute each solution back into the original equation.

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

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: The equation is in the correct form already.

Factor: Rewrite $x^2 - 12x + 36 =0$ as $x^2 - 2 \cdot (-6)x + ( -6)^2$.

We recognize this as a perfect square. This factors as $(x - 6)^2 = 0$ or $(x - 6)(x - 6) = 0$

Set each factor equal to zero:

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

Solve:

$\underline{\underline{x = 6}} && \text{or} && \underline{\underline{x = 6}}$

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.

$x = 6 \quad \quad 6^2 - 12(6) + 36 = 36 - 72 + 36 = 0 \quad \quad \text{checks out}$

b) $x^2 - 81 = 0$

Rewrite: this is not necessary since the equation is in the correct form already

Factor: Rewrite $x^2 - 81$ as $x^2 - 9^2$.

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

Set each factor equal to zero:

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

Solve:

$\underline{\underline{x = 9}} && \text{or} && \underline{\underline{x = - 9}}$

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 $x^2 + 20x + 100$ 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:

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

Solve:

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

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 differences 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. $4x^2 - y^2$
9. $16x^2 - 81y^2$

Solve the following quadratic equations 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. $-x^2 - 16x - 60 = 0$

Feb 23, 2012

## Last Modified:

Sep 15, 2014
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