Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set - Page 345: 95



Work Step by Step

Grouping the first $3$ terms the given expression is equivalent to \begin{array}{l}\require{cancel} 36-12x+x^2-a^2 \\\\= (36-12x+x^2)-a^2 \\\\= (x^2-12x+36)-a^2 .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} (x^2-12x+36) \end{array} has $c= 36 $ and $b= -12 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -6,-6 \right\}.$ Using these two numbers, the $\text{ expression }$ above is equivalent to \begin{array}{l}\require{cancel} (x-6)(x-6)-a^2 \\\\= (x-6)^2-a^2 .\end{array} The expressions $ (x-6)^2 $ and $ a^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ (x-6)^2-a^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x-6)^2-(a)^2 \\\\= (x-6+a)(x-6-a) .\end{array}
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