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 R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set: 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}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.