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: 39

Answer

$(2r+5s)(5r-s)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 10r^2+23rs-5s^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 10 ,b= 23 ,\text{ and } c= -5 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 10(-5)=-50 $ and whose sum is $b$ are $\left\{ 25,-2 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 10r^2+25rs-2rs-5s^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (10r^2+25rs)-(2rs+5s^2) \\\\= 5r(2r+5s)-s(2r+5s) \\\\= (2r+5s)(5r-s) .\end{array}
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