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# 9.6: Factoring Special Products

Difficulty Level: At Grade Created by: CK-12

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 lesson, 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: a2b2\begin{align*}a^2-b^2\end{align*}. 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.

a2b2=(a+b)(ab)

In these problems, the key is figuring out what the a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} terms are. Let’s do some examples of this type.

Example 1: Factor the difference of squares.

(a) x29\begin{align*}x^2-9\end{align*}

(b) x2y21\begin{align*}x^2y^2-1\end{align*}

Solution:

(a) Rewrite as x29\begin{align*}x^2-9\end{align*} as x232\begin{align*}x^2-3^2\end{align*}. Now it is obvious that it is a difference of squares.

We substitute the values of a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} for the Sum and Difference Formula:

(x+3)(x3)

The answer is x29=(x+3)(x3)\begin{align*}x^2-9=(x+3)(x-3)\end{align*}.

(b) Rewrite as x2y21\begin{align*}x^2y^2-1\end{align*} as (xy)212\begin{align*}(xy)^2-1^2\end{align*}. This factors as (xy+1)(xy1)\begin{align*}(xy+1)(xy-1)\end{align*}.

## Factoring Perfect Square Trinomials

A perfect square trinomial has the form

a2+2ab+b2ora22ab+b2

The factored form of a perfect square trinomial has the form

And(a+b)2 if a2+2(ab)+b2(ab)2 if a22(ab)+b2

In these problems, the key is figuring out what the a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} terms are. Let’s do some examples of this type.

Example: x2+8x+16\begin{align*}x^2+8x+16\end{align*}

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

x2+8x+16asx2+8x+42.

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.

x2+8x+16asx2+24x+42

This means we can factor x2+8x+16\begin{align*}x^2+8x+16\end{align*} as (x+4)2\begin{align*}(x+4)^2\end{align*}.

Example 2: Factor x24x+4\begin{align*}x^2-4x+4\end{align*}.

Solution: Rewrite x24x+4\begin{align*}x^2-4x+4\end{align*} as x2+2(2)x+(2)2\begin{align*}x^2+2 \cdot (-2) \cdot x+(-2)^2\end{align*}.

We notice that this is a perfect square trinomial and we can factor it as: (x2)2\begin{align*}(x-2)^2\end{align*}.

## Solving Polynomial Equations Involving Special Products

We have learned how to factor quadratic polynomials that are helpful in solving polynomial equations like ax2+bx+c=0\begin{align*}ax^2+bx+c=0\end{align*}. 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.

Example 3: Solve the following polynomial equations.

x2+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:

6x2+7x+6=6×1=0andfactors as6+1=7(x+1)(x+6)=0

Set each factor equal to zero:

x+1=0orx+6=0

Solve:

x=1orx=6

Check: Substitute each solution back into the original equation.

(1)2+7(1)+6(6)2+7(6)+6=1+(7)+6=0=36+(42)+6=0

## Practice Set

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.

Factor the following perfect square trinomials.

1. x2+8x+16\begin{align*}x^2+8x+16\end{align*}
2. x218x+81\begin{align*}x^2-18x+81\end{align*}
3. x2+24x144\begin{align*}-x^2+24x-144\end{align*}
4. x2+14x+49\begin{align*}x^2+14x+49\end{align*}
5. 4x24x+1\begin{align*}4x^2-4x+1\end{align*}
6. 25x2+60x+36\begin{align*}25x^2+60x+36\end{align*}
7. 4x212xy+9y2\begin{align*}4x^2-12xy+9y^2\end{align*}
8. x4+22x2+121\begin{align*}x^4+22x^2+121\end{align*}

Factor the following difference of squares.

1. x24\begin{align*}x^2-4\end{align*}
2. x236\begin{align*}x^2-36\end{align*}
3. x2+100\begin{align*}-x^2+100\end{align*}
4. x2400\begin{align*}x^2-400\end{align*}
5. 9x24\begin{align*}9x^2-4\end{align*}
6. 25x249\begin{align*}25x^2-49\end{align*}
7. 36x2+25\begin{align*}-36x^2+25\end{align*}
8. 16x281y2\begin{align*}16x^2-81y^2\end{align*}

Solve the following quadratic equations using factoring.

1. x211x+30=0\begin{align*}x^2-11x+30=0\end{align*}
2. x2+4x=21\begin{align*}x^2+4x=21\end{align*}
3. x2+49=14x\begin{align*}x^2+49=14x\end{align*}
4. x264=0\begin{align*}x^2-64=0\end{align*}
5. x224x+144=0\begin{align*}x^2-24x+144=0\end{align*}
6. 4x225=0\begin{align*}4x^2-25=0\end{align*}
7. x2+26x=169\begin{align*}x^2+26x=-169\end{align*}
8. x216x60=0\begin{align*}-x^2-16x-60=0\end{align*}

Mixed Review

1. Find the value for k\begin{align*}k\end{align*} that creates an infinite number of solutions to the system {3x+7y=1kx14y=2\begin{align*}\begin{cases} 3x+7y=1\\ kx-14y=-2 \end{cases}\end{align*}.
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 y5=13(x+4)\begin{align*}y-5= \frac{1}{3}(x+4)\end{align*}. 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?
5. Divide 489÷315\begin{align*}4 \frac{8}{9} \div -3\frac{1}{5}\end{align*}.
6. Identify an integer than is even and not a natural number.

8 , 9

## Date Created:

Feb 22, 2012

Dec 11, 2014
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