Factor $x^2+x+1+i$ in $\mathbb{C}[x]$.

Question

Factor $x^2+x+1+i$ in $\mathbb{C}[x]$.

I know the roots are $-i$ and $-1+i$, but I don't know how to go about factoring such polynomial. I tried using the quadratic formula, but I got stuck half way.

Edit: Obviously, I know once I have the roots how to factor it. I am asking how can I find the roots rigorously, and not by plugging in values hoping to find a root.

Answer 1

Hint for quadratic . We get $-1+(-3-4i)^{1/2}/2$ and same with minus sign so two roots are conjugate. Can you find the square root of $-3-4i$. Using the relation let $(-3-4i)^{1/2}=a+bi$ so squaring and comparing we get $a^2-b^2=-3$ and $2ab=-4$ . Find $a$ and $b$ where $a$ is the real part while $b$ is the imaginary part of square root of $-3-4i$, plug this obtained value in place of $-3-4i$ in the quadratic formula and you get roots; once you have roots you can factorize.

Answer 2

If a degree $2$ polynomial $p(x)=ax^2+bx+c$ has roots $x_1$ and $x_2$, then: $$p(x)=a(x-x_1)(x-x_2).$$

Hence, $$x^2 + x + 1 + i = (x+i)(x-i+1).$$

EDIT: You just use the quadratic formula as per normal: $$x=\frac{-1\pm\sqrt{1-4(1+i)}}{2}=\frac{-1\pm \sqrt{-3-4i}}{2}.$$ To find $\sqrt{-3-4i}$, you can convert to polar form, or do it algebraically.