Guessing root of polynomials

Question

Given $p(x)=x^5+(1+2i)x^4-(1+3i)x^2+8+44i$ check with the Horner-scheme if $(-2-i)$ is a root of $p(x)$.

First I have to guess a root, then proceed with the Horner-method and if i factorized it, i can say if $(-2-i)$ is a root or not, but how can i guess the first root, are there any tricks ?

Answer 1

$p(x) = 8 + 44i + x^2(-(1+3i) + x^2(1+2i + x))$

Now, let $x=-2-i$, thus, $x^2 = 4-1-4i = 3+4i$

$\begin{align}p(x) &= 8 +44i + (3+4i)(-(1+3i) + (3+4i)(-1+i)))\\ &=8+44i + (3+4i)(-(1+3i) -3+3i-4i-4))\\ &=8+44i +(3+4i)(-1-3i-3+3i-4i-4)\\ &=8+44i + (3+4i)(-8 - 4i)\\ &=8+44i -24 -12i -32i +16\\ &=0 \end{align} $

Thus $-2-i$ is a root.

Answer 2

You don't need to factor the entire thing. $r$ is a root of $p(x)$ if and onyl if $p(r)=0$. So use Horner's scheme to evaluate $p(-2-i)$ and see if you get $0$ or not.