## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Review Exercises: 33

#### Answer

$(x^4+1)(x^2+1)(x+1)(x-1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $x^8-1 ,$ use the factoring of the difference of $2$ squares repetitively until none of the resulting expressions is still a difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $x^8$ and $1$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^8-1 ,$ 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} (x^4)^2-(1)^2 \\\\= (x^4+1)(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} (x^4+1)[(x^2)^2-(1)^2] \\\\= (x^4+1)[(x^2+1)(x^2-1)] \\\\= (x^4+1)(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} (x^4+1)(x^2+1)[(x)^2-(1)^2] \\\\= (x^4+1)(x^2+1)[(x+1)(x-1)] \\\\= (x^4+1)(x^2+1)(x+1)(x-1) .\end{array}

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