Question

Factor ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{+}\mathit{i}$ in $\mathbf{\square }\left[x\right]$.

Short answer

The factors of $x^2+x+1+i$ is $\left(x+i\right)\left(x+1-i\right)$.

Step-by-step solution

Polynomial in □x

Every non $-$constant polynomial in $\square \left[x\right]$has a root in $\square$.

Factors of  x2+x+1+i

Given that the polynomial $f\left(x\right)=x^2+x+1+i$. Rewrite the polynomial as $f\left(x\right)=x^2+ix+x-ix+1+i$ and simplify.

$$\begin{gathered} x^2+x+1+i=x^2+ix+x-ix+1+i \\ =x\left(x+1-i\right)+i\left(x+1-i\right) \\ =\left(x+1-i\right)\left(x-i\right) \end{gathered}$$

Therefore, the factors of $f\left(x\right)=x^2+x+1+i$ are $\left(x+1-i\right)\left(x+i\right)$.