#### 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}