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: 58

Answer

$(p-q+5)(p^2-2pq-q^2-5p-5q+25)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ (p-q)^3+125 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using the factoring of the sum/difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2),$ the expression above is equivalent to \begin{array}{l}\require{cancel} [(p-q)+5][(p-q)^2-(p-q)(5)+(5)^2] \\\\= (p-q+5)[(p-q)^2-5p-5q+25] .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to\begin{array}{l}\require{cancel} (p-q+5)[(p^2-2pq-q^2)-5p-5q+25] \\\\= (p-q+5)(p^2-2pq-q^2-5p-5q+25) .\end{array}
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