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 347: 3

Answer

$3p^2(p-6)(p+5)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 3p^4-3p^3-90p^2 ,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $x^2+bx+c.$ $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $ 3p^2 ,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 3p^2(p^2-p-30) .\end{array} In the trinomial expression above, $a= 1 ,b= -1 ,\text{ and } c= -30 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ -6,5 \right\}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} 3p^2(p-6)(p+5) .\end{array}
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