# Solve the equation $x^{4}-8x^{3}+24x^{2}-32x+20$ = 0 if $3+\mathit{i}$ is a root.

Toolbox:
• In a polynomial equation with real coefficients,complex roots occur in conjugate pairs.
Step 1:
$x^4-8x^3+24x^2-32x+20=0$
One of the roots of this equation is $3+i$
Therefore another root is $3-i$
Therefore $[x-(3+i)][x-(3-i)]$ is a quadratic factor of the LHS.
$\Rightarrow [(x-3)-i][(x-3)+i]=(x-3)^2+1$
$\qquad\qquad\qquad\qquad\qquad\;\;\;=x^2-6x+9+1$
$\qquad\qquad\qquad\qquad\qquad\;\;\;=x^2-6x+10$
Step 2:
Therefore $x^4-8x^3+24x^2-32x+20=(x^2-6x+10)(x^2+px+q)$
It can be seen $q=2$ (equating the constants)
Equating the coefficient of $x$
$-32=-12+10p$
$10p=-20\Rightarrow p=-2$
Therefore the other quadratic factor is $x^2-2x+2$.
Step 3:
The other two roots are obtained by solving $x^2-2x+2=0$
$x=\large\frac{2\pm\sqrt{4-8}}{2}$
$\;\;=\large\frac{2\pm 2i}{2}$
$\;\;=1\pm i$
The roots of the equations are $3\pm i,1\pm i$
edited Jun 11, 2013