## Algebra 2 Common Core

$(x+i)(x-i)$
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*}