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: 53

Answer

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

Work Step by Step

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