Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises: 54

Answer

$5(w^{8}+1)(w^{4}+1)(w^{2}+1)(w+1)(w-1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 5w^{16}-5 ,$ factor first the $GCF.$ Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Factoring the $GCF=5,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 5(w^{16}-1) .\end{array} 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} 5[(w^{8})^2-(1)^2] \\\\= 5(w^{8}+1)(w^{8}-1) \\\\= 5(w^{8}+1)[(w^{4})^2-(1)^2] \\\\= 5(w^{8}+1)(w^{4}+1)(w^{4}-1) \\\\= 5(w^{8}+1)(w^{4}+1)[(w^{2})^2-(1)^2] \\\\= 5(w^{8}+1)(w^{4}+1)(w^{2}+1)(w^{2}-1) \\\\= 5(w^{8}+1)(w^{4}+1)(w^{2}+1)[(w)^{2}-(1)^2] \\\\= 5(w^{8}+1)(w^{4}+1)(w^{2}+1)(w+1)(w-1) .\end{array}
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