# 6.5 Factoring two special products

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses.The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: know the fundamental rules of factoring, be able to factor the difference of two squares and perfect square trinomials.

## Overview

• The Difference of Two Squares
• Fundamental Rules of Factoring
• Perfect Square Trinomials

## The difference of two squares

Recall that when we multiplied together the two binomials $\left(a+b\right)$ and $\left(a-b\right)$ , we obtained the product ${a}^{2}-{b}^{2}$ .

$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$

## Perfect square

Notice that the terms ${a}^{2}$ and ${b}^{2}$ in the product can be produced by squaring $a$ and $b$ , respectively. A term that is the square of another term is called a perfect square . Thus, both ${a}^{2}$ and ${b}^{2}$ are perfect squares. The minus sign between ${a}^{2}$ and ${b}^{2}$ means that we are taking the difference of the two squares.
Since we know that $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ , we need only turn the equation around to find the factorization form.

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

The factorization form says that we can factor ${a}^{2}-{b}^{2}$ , the difference of two squares, by finding the terms that produce the perfect squares and substituting these quantities into the factorization form.
When using real numbers (as we are), there is no factored form for the sum of two squares. That is, using real numbers,

${a}^{2}+{b}^{2}$ cannot be factored

## Sample set a

Factor ${x}^{2}-16$ . Both ${x}^{2}$ and 16 are perfect squares. The terms that, when squared, produce ${x}^{2}$ and 16 are $x$ and 4, respectively. Thus,

${x}^{2}-16=\left(x+4\right)\left(x-4\right)$

We can check our factorization simply by multiplying.

$\begin{array}{lll}\left(x+4\right)\left(x-4\right)\hfill & =\hfill & {x}^{2}-4x+4x-16\hfill \\ \hfill & =\hfill & {x}^{2}-16.\hfill \end{array}$

$49{a}^{2}{b}^{4}-121$ . Both $49{a}^{2}{b}^{4}$ and 121 are perfect squares. The terms that, when squared, produce $49{a}^{2}{b}^{4}$ and 121 are $7a{b}^{2}$ and 11, respectively. Substituting these terms into the factorization form we get

$49{a}^{2}{b}^{4}-121=\mathrm{\left(7}a{b}^{2}+11\right)\left(7a{b}^{2}-11\right)$

We can check our factorization by multiplying.

$\begin{array}{lll}\left(7a{b}^{2}+11\right)\left(7a{b}^{2}-11\right)\hfill & =\hfill & 49{a}^{2}{b}^{4}-11a{b}^{2}+11a{b}^{2}-121\hfill \\ \hfill & =\hfill & 49{a}^{2}{b}^{4}-121\hfill \end{array}$

$3{x}^{2}-27$ . This doesn’t look like the difference of two squares since we don’t readily know the terms that produce $3{x}^{2}$ and 27. However, notice that 3 is common to both the terms. Factor out 3.

$3\left({x}^{2}-9\right)$

Now we see that ${x}^{2}-9$ is the difference of two squares. Factoring the ${x}^{2}-9$ we get

$\begin{array}{lll}3{x}^{2}-27\hfill & =\hfill & 3\left({x}^{2}-9\right)\hfill \\ \hfill & =\hfill & 3\left(x+3\right)\left(x-3\right)\hfill \end{array}$

Be careful not to drop the factor 3.

## Practice set a

If possible, factor the following binomials completely.

${m}^{2}-25$

$\left(m+5\right)\left(m-5\right)$

$36{p}^{2}-81{q}^{2}$

$9\left(2p-3q\right)\left(2p+3q\right)$

$49{a}^{4}-{b}^{2}{c}^{2}$

$\left(7{a}^{2}+bc\right)\left(7{a}^{2}-bc\right)$

${x}^{8}{y}^{4}-100{w}^{12}$

$\left({x}^{4}{y}^{2}+10{w}^{6}\right)\left({x}^{4}{y}^{2}-10{w}^{6}\right)$

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