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 - Mid-Chapter Review - Mixed Review: 9

Answer

$6(p+t)(p-t)$

Work Step by Step

Factoring the $GCF= 6 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 6p^2-6t^2 \\\\= 6(p^2-t^2) .\end{array} The expressions $ p^2 $ and $ t^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ p^2-t^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6[(p)^2-(t)^2] \\\\= 6[(p+t)(p-t)] \\\\= 6(p+t)(p-t) .\end{array}
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