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 4 - Polynomials - 4.6 Special Products - 4.6 Exercise Set - Page 277: 64



Work Step by Step

Using $a(b+c)=ab+ac$ or the Distributive Property, the given expression, $ (x^2-5)(x^2+x-1) $, is equivalent to \begin{array}{l}\require{cancel} x^2(x^2)+x^2(x)+x^2(-1)-5(x^2)-5(x)-5(-1) \\\\= x^4+x^3-x^2-5x^2-5x+5 \\\\= x^4+x^3+(-x^2-5x^2)-5x+5 \\\\= x^4+x^3-6x^2-5x+5 .\end{array}
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