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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.

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    $\begingroup$ If those are the roots... then $(x+i)(x-(-1+i))$? $\endgroup$ – Euler....IS_ALIVE Nov 17 '15 at 6:17
  • $\begingroup$ $x^2 + x + 1 + i = (x+i)(x-i+1)$ $\endgroup$ – user171326 Nov 17 '15 at 6:18
  • $\begingroup$ @Euler....IS_ALIVE check edit $\endgroup$ – Chad Nov 17 '15 at 6:40
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    $\begingroup$ @ Chad I have answered your edit $\endgroup$ – Archis Welankar Nov 17 '15 at 7:06
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    $\begingroup$ You can use the quadratic formula to find the roots, as you would in $\mathbb R$. Here $a = 1$, $b = 1$, $c = 1+i$. So we have $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-1 \pm \sqrt{-3-4i}}{2} = \frac{-1 \pm (1-2i)}{2} = \{ -i, -1 + i \}$$ $\endgroup$ – amcerbu Nov 17 '15 at 7:18
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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.

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  • $\begingroup$ Thank you for the detailed explanation! I edited what is inside the radical, which should be $-3-4i$ not $-3+4i$ as you had it. Thanks, again! $\endgroup$ – Chad Nov 19 '15 at 2:57
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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.

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  • $\begingroup$ A "monic" polynomial yes. I general we have the coeffient of $x^2$ as a factor as well, e.g. $\endgroup$ – Henno Brandsma Nov 17 '15 at 6:19
  • $\begingroup$ Yes, this is correct. $\endgroup$ – MrMazgari Nov 17 '15 at 6:24
  • $\begingroup$ check the edit made $\endgroup$ – Chad Nov 17 '15 at 6:41
  • $\begingroup$ The first sentence is still not correct $\endgroup$ – Henno Brandsma Nov 17 '15 at 11:02

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