Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises - Page 348: 63

Answer

$(p+8q-5)^2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ (p+8q)^2-10(p+8q)+25 ,$ use substitution. Then use the factoring of trinomials in the form $x^2+bx+c.$ $\bf{\text{Solution Details:}}$ Let $z=(p+8q).$ Then the given expression is equivalent to \begin{array}{l}\require{cancel} z^2-10z+25 .\end{array} In the trinomial expression above, $b= -10 ,\text{ and } c= 25 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ -5,-5 \right\}.$ Using these two numbers, the factored form of the expression above is \begin{array}{l}\require{cancel} (z-5)(z-5) \\\\= (z-5)^2 .\end{array} Since $z=(p+8q),$ by back substitution, the expression above is equivalent to \begin{array}{l}\require{cancel} (p+8q-5)^2 .\end{array}
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