Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 33

Answer

$(p+q+1)^2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ (p+q)^2+2(p+q)+1 ,$ simplify the expression using substitution. The resulting expression is a perfect square trinomial. Use the factoring of perfect square trinomials. Finally, substitute back the original expression. $\bf{\text{Solution Details:}}$ Let $z= (p+q) .$ The given expression becomes \begin{array}{l}\require{cancel} z^2+2z+1 .\end{array} The trinomial above is a perfect square trinomial. Using $a^2\pm2ab+b^2=(a\pm b)^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (z+1)^2 .\end{array} Since $z= (p+q) ,$, then the expression above becomes \begin{array}{l}\require{cancel} ((p+q)+1)^2 \\\\ (p+q+1)^2 .\end{array}
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