Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set: 95

Answer

$-(3x^{m}-4)(5x^m-2)$

Work Step by Step

Factoring the negative $GCF= -1 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -15x^{2m}+26x^m-8 \\\\= -(15x^{2m}-26x^m+8) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -(15x^{2m}-26x^m+8) \end{array} has $ac= 15(8)=120 $ and $b= -26 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -20,-6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -(15x^{2m}-20x^m-6x^m+8) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -[(15x^{2m}-20x^m)-(6x^m-8)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -[5x^m(3x^{m}-4)-2(3x^m-4)] .\end{array} Factoring the $GCF= (3x^{m}-4) $ of the entire expression above results to \begin{array}{l}\require{cancel} -[(3x^{m}-4)(5x^m-2)] \\\\= -(3x^{m}-4)(5x^m-2) .\end{array}
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