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