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: 33

Answer

$6z(2z^2-z+3)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 12z^3-6z^2+18z ,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $ 6z ,$ 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} 6z(2z^2-z+3) .\end{array} In the trinomial expression above, $a= 2 ,b= -1 ,\text{ and } c= 3 .$ There are no two numbers whose product is $ac= 2(3)=6 $ and whose sum is $b.$ Hence, the factored form is \begin{array}{l}\require{cancel} 6z(2z^2-z+3) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.