Question

Solve the equation $x^4-8x^3+24x^2-32x+20=0$ if $3+i$ is a root of it.

Answer 1

Since $3+i$ is a root, $3-i$ is another root.
Sum of the roots $=3+i+3-i=6$
Product of the roots $=(3+i)(3-i)=3^2+1^2=10$
The equation with $3+i$ and $3-i$ as roots is
$x^2-\left(6\right)x+10=0$
(i.e.), $x^2-6x+10=0$
$\Rightarrow x^4-8x^3+24x^2-32x+20=(x^2-6x+10)(x^2+px+2)$
Equating co-efficient of $x$
$\Rightarrow -12+10p=-32$
$\Rightarrow 10p=-20$
$\Rightarrow p=-2$
$\therefore$ the other factor is $x^2-2x+2$
Solving $x^2-2x+2=0$, we get

$a=1$, $b=-2$, $c=2$
Therefore, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{+2\pm \sqrt{4-8}}{2}$

$=\frac{2\pm 2i}{2}$

$=\frac{2(1\pm i)}{2}$
$=1\pm i$