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: 21

Answer

$\left(x^2+\dfrac{1}{4} \right)\left(x+\dfrac{1}{2} \right)\left(x-\dfrac{1}{2} \right)$

Work Step by Step

The given expression is equivalent to \begin{array}{l}\require{cancel} -\dfrac{1}{16}+x^4 \\\\= x^4-\dfrac{1}{16} .\end{array} The expressions $ x^4 $ and $ \dfrac{1}{16} $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^4-\dfrac{1}{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} (x^2)^2-\left( \dfrac{1}{4} \right)^2 \\\\= \left(x^2+\dfrac{1}{4} \right)\left(x^2-\dfrac{1}{4} \right) .\end{array} The expressions $ x^2 $ and $ \dfrac{1}{4} $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^2-\dfrac{1}{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} \left(x^2+\dfrac{1}{4} \right)\left(x^2-\dfrac{1}{4} \right) \\\\= \left(x^2+\dfrac{1}{4} \right)\left[ \left(x\right)^2-\left(\dfrac{1}{2} \right)^2\right] \\\\= \left(x^2+\dfrac{1}{4} \right)\left(x+\dfrac{1}{2} \right)\left(x-\dfrac{1}{2} \right) .\end{array}
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