Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 70: 124


In order to factor by grouping, we arrange the terms in pairs and factor out the common factor for each pair, hoping to obtain an expression of the form =(term 1)[expression 1]+(term 2)[expression 1], that is, the brackets hold $\underline{the~same~expression}$. If that is the case, we factor it out: =[expression 1][(term 1) + (term 2) ] --- For example, $3x^{2}+3y^{2}x+2y^{2}x+2y^{4}=[3x^{2}+3y^{2}x]+[2y^{2}x+2y^{4}]$ $=3x(x+y^{2})+2y^{2}(x+y^{2})\quad $...same expression! =$(x+y^{2})(3x+2y^{2})$

Work Step by Step

Given above.
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.