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