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 - 5.6 Factoring: A General Strategy - 5.6 Exercise Set: 70

Answer

$a^2b(ab+5)(ab-3)$

Work Step by Step

Factoring the $GCF= a^2b $, then the given expression, $ a^4b^3+2a^3b^2-15a^2b $ is equivalent to \begin{array}{l} a^2b(a^2b^2+2ab-15) .\end{array} The two numbers whose product is $ac= 1(-15)=-15 $ and whose sum is $b= 2 $ are $\{ 5,-3 \}$. Using these two numbers to decompose the middle term of the expression, $ a^2b(a^2b^2+2ab-15) ,$ then the factored form is \begin{array}{l} a^2b(a^2b^2+5ab-3ab-15) \\\\= a^2b[(a^2b^2+5ab)-(3ab+15)] \\\\= a^2b[ab(ab+5)-3(ab+5)] \\\\= a^2b[(ab+5)(ab-3)] \\\\= a^2b(ab+5)(ab-3) .\end{array}
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