Answer
$\left(x+\dfrac{1}{2}i\right)\left(x-\dfrac{1}{2}i\right)$
Work Step by Step
The given expression, $ x^2+\dfrac{1}{4} ,$ is equivalent to \begin{align*} & x^2-\left(-\dfrac{1}{4}\right) \\\\&= x^2-\left(-1\cdot\dfrac{1}{4}\right) .\end{align*} Since $i^2=-1,$ the expression above is equivalent to \begin{align*} & x^2-\left(i^2\cdot\dfrac{1}{4}\right) \\\\&= x^2-\left(\dfrac{1}{4}i^2\right) .\end{align*} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the expression above is equivalent to \begin{align*} & (x)^2-\left(\dfrac{1}{2}i\right)^2 \\&= \left(x+\dfrac{1}{2}i\right)\left(x-\dfrac{1}{2}i\right) .\end{align*} Hence, the factored form of the given expression is $
\left(x+\dfrac{1}{2}i\right)\left(x-\dfrac{1}{2}i\right)
.$
Checking: Multiplying the factors above results to
\begin{align*}
&
\left(x+\dfrac{1}{2}i\right)\left(x-\dfrac{1}{2}i\right)
\\\\&=
\left(x\right)^2-\left(\dfrac{1}{2}i\right)^2
&\left(\text{use }(a+b)(a-b)=a^2-b^2\right)
\\\\&=
x^2-\dfrac{1}{4}i^2
\\\\&=
x^2-\dfrac{1}{4}(-1)
&\left(\text{use }i^2=-1\right)
\\\\&=
x^2+\dfrac{1}{4}
\text{ (same as original expression)}
.\end{align*}
