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: 69

Answer

$5(pt+6)(pt-1)$

Work Step by Step

Factoring the $GCF= 5 $, then the given expression, $ 5p^2t^2+25pt-30 $ is equivalent to \begin{array}{l} 5(p^2t^2+5pt-6) .\end{array} The two numbers whose product is $ac= 1(-6)=-6 $ and whose sum is $b= 5 $ are $\{ 6,-1 \}$. Using these two numbers to decompose the middle term of the expression, $ 5(p^2t^2+5pt-6) ,$ then the factored form is \begin{array}{l} 5(p^2t^2+6pt-1pt-6) \\\\= 5[(p^2t^2+6pt)-(1pt+6)] \\\\= 5[pt(pt+6)-(pt+6)] \\\\= 5[(pt+6)(pt-1)] \\\\= 5(pt+6)(pt-1) .\end{array}
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