Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 70

Answer

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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 1-x^{16} ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 1 $ and $ x^{16} $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 1-x^{16} ,$ 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} (1)^2-(x^{8})^2 \\\\= (1+x^{8})(1-x^{8}) .\end{array} The expressions $ 1 $ and $ x^{8} $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 1-x^{8} ,$ 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} (1+x^{8})[(1)^2-(x^{4})^2] \\\\= (1+x^{8})[(1+x^{4})(1-x^{4})] \\\\= (1+x^{8})(1+x^{4})(1-x^{4}) .\end{array} The expressions $ 1 $ and $ x^{4} $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 1-x^{4} ,$ 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} (1+x^{8})(1+x^{4})[(1)^2-(x^{2})^2] \\\\ (1+x^{8})(1+x^{4})[(1+x^{2})(1-x^{2})] \\\\ (1+x^{8})(1+x^{4})(1+x^{2})(1-x^{2}) .\end{array} The expressions $ 1 $ and $ x^{2} $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 1-x^{2} ,$ 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} (1+x^{8})(1+x^{4})(1+x^{2})[(1)^2-(x)^{2}] \\\\= (1+x^{8})(1+x^{4})(1+x^{2})[(1+x)(1-x)] \\\\= (1+x^{8})(1+x^{4})(1+x^{2})(1+x)(1-x) .\end{array}
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