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# 9.9: Factor Polynomials Using Special Products

Difficulty Level: Basic Created by: CK-12
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Practice Factor Polynomials Using Special Products
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What if the area of a square playground were 10,000 square feet? Instead of taking the square root of the area to find the length of one of the playground's sides, you could set up the equation $s^2 = 10,000$ , subtract 10,000 from both sides to get $s^2 - 10,000 = 0$ , and solve for $s$ by factoring $s^2 -10,000$ on the left side of the equation. But how would you factor this? After completing this Concept, you'll know how to factor expressions like this one by using special products.

### Guidance

When we learned how to multiply binomials, we talked about two special products: the Sum and Difference Formula and the Square of a Binomial Formula. In this Concept, we will learn how to recognize and factor these special products.

Factoring 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 A

Factor the difference of squares.

(a) $x^2-9$

(b) $x^2y^2-1$

Solution:

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

We substitute the values of $a$ and $b$ in the Sum and Difference Formula:

$(x+3)(x-3)$

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

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

Factoring Perfect Square Trinomials

A perfect square trinomial has the form:

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

The factored form of a perfect square trinomial has the form:

$&&(a+b)^2 \ if \ a^2+2(ab)+b^2\\\text{And}\\&&(a-b)^2 \ if \ a^2-2(ab)+b^2$

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

#### Example B

Factor $x^2+8x+16$ .

Solution:

Check that the first term and the last term are perfect squares.

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

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.

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

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

#### Example C

Factor $x^2-4x+4$ .

Solution:

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 $(x-2)^2$ .

Solving Polynomial Equations Involving Special Products

We have learned how to factor quadratic polynomials that are helpful in solving polynomial equations like $ax^2+bx+c=0$ . Remember that to solve polynomials in expanded form, we use the following steps:

Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0.

Step 2: Factor the polynomial completely.

Step 3: Use the Zero Product Property to set each factor equal to zero.

Step 4: Solve each equation from step 3.

### Guided Practice

Solve the following polynomial equations.

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

Solution: No need to rewrite because it is already in the correct form.

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

$6& =6 \times 1 && \text{and} && 6+1=7\\x^2+7x+6 & = 0 && \text{factors as} && (x+1)(x+6)=0$

Set each factor equal to zero:

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

Solve:

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

Check: Substitute each solution back into the original equation.

$(-1)^2+7(-1)+6&=1+(-7)+6=0\\(-6)^2+7(-6)+6&=36+(-42)+6=0$

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Factoring Special Products (10:08)

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. $16x^2-81y^2$

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. $-x^2-16x-60=0$

Mixed Review

1. Find the value for $k$ that creates an infinite number of solutions to the system $\begin{cases}3x+7y=1\\kx-14y=-2 \end{cases}$ .
2. A restaurant has two kinds of rice, three choices of mein, and four kinds of sauce. How many plate combinations can be created if you choose one of each?
3. Graph $y-5= \frac{1}{3}(x+4)$ . Identify its slope.
4. \$600 was deposited into an account earning 8% interest compounded annually.
1. Write the exponential model to represent this situation.
2. How much money will be in the account after six years?
1. Divide: $4 \frac{8}{9} \div -3\frac{1}{5}$ .
2. Identify an integer than is even and not a natural number.

### Vocabulary Language: English Spanish

difference of two squares

difference of two squares

The difference of two squares has the form $a^2-b^2=(a+b)(a-b)$.
Difference of Squares

Difference of Squares

A difference of squares is a quadratic equation in the form $a^2-b^2$.
Perfect Square Trinomial

Perfect Square Trinomial

A perfect square trinomial is a quadratic expression of the form $a^2+2ab+b^2$ (which can be rewritten as $(a+b)^2$) or $a^2-2ab+b^2$ (which can be rewritten as $(a-b)^2$).

A polynomial in quadratic form looks like a trinomial or binomial and can be factored like a quadratic expression.

Basic

8 , 9

Feb 24, 2012

Feb 26, 2015