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 5 - Polynomials and Factoring - Mid-Chapter Review - Mixed Review: 13

Answer

$10(x^2+1)(x+1)(x-1)$

Work Step by Step

Factoring the $GCF= 10 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 10x^4-10 \\\\= 10(x^4-1) .\end{array} The expressions $ x^4 $ and $ 1 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^4-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 10(x^4-1) \\\\= 10[(x^2)^2-(1)^2] \\\\= 10(x^2+1)(x^2-1) .\end{array} The expressions $ x^2 $ and $ 1 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^2-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 10(x^2+1)(x^2-1) \\\\= 10(x^2+1)[(x)^2-(1)^2] \\\\= 10(x^2+1)[(x+1)(x-1)] \\\\= 10(x^2+1)(x+1)(x-1) .\end{array}
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