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: 28

Answer

$(m-3n)(2m+5n)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 2m^2-mn-15n^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 2 ,b= -1 ,\text{ and } c= -15 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 2(-15)=-30 $ and whose sum is $b$ are $\left\{ -6,5 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 2m^2-6mn+5mn-15n^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (2m^2-6mn)+(5mn-15n^2) \\\\= 2m(m-3n)+5n(m-3n) \\\\= (m-3n)(2m+5n) .\end{array}
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