Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.3 - Multiplying and Factoring Algebraic Expressions - Exercises: 21

Answer

$y^{5}+4y^{4} +4y^{3}-y$

Work Step by Step

If we factor out y from the first parentheses, $y(y^{2}+2y+1)(y^{2}+2y-1)=...,$ we may recognize a special product, a difference of squares: ... $I^{2}-II^{2}=(I+II)(I-II)$ $=y[(y^{2}+2y)+1]\cdot[(y^{2}+2y)-1]=y[(y^{2}+2y)^{2}-1^{2}]$ ... first term in the brackets is a square of a sum ... ... $(I+II)^{2}= I^{2}+2\cdot I\cdot II+II^{2}$, $=y[(y^{2})^{2}+2(y^{2})(2y)+(2y)^{2}-1]$ $= y(y^{4}+4y^{3}+4y^{2}-1)$ $=y^{5}+4y^{4} +4y^{3}-y$
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.