## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 22

#### Answer

$5(x^4+z^{8})(x^2+z^{4})(x+z^{2})(x-z^{2})$

#### Work Step by Step

Factoring the $GCF=5,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 5x^8-5z^{16} \\\\= 5(x^8-z^{16}) .\end{array} The expressions $x^8$ and $z^{16}$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^8-z^{16} ,$ 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} 5(x^8-z^{16}) \\\\= 5[(x^4)^2-(z^{8})^2] \\\\= 5[(x^4+z^{8})(x^4-z^{8})] \\\\= 5(x^4+z^{8})(x^4-z^{8}) \\\\= 5(x^4+z^{8})[(x^2)^2-(z^{4})^2] \\\\= 5(x^4+z^{8})(x^2+z^{4})(x^2-z^{4}) \\\\= 5(x^4+z^{8})(x^2+z^{4})[(x)^2-(z^{2})^2] \\\\= 5(x^4+z^{8})(x^2+z^{4})(x+z^{2})(x-z^{2}) .\end{array}

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