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

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