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# Solve the equation $x^{4}-4x^{3}+11x^{2}-14x+10=0$ if one root is $1+2i$.

Toolbox:
• In a polynomial equation with real coefficients,complex roots occur in conjugate pairs.
Step 1:
$x^4-4x^3+11x^2-14x+10=0$
One of the roots of this equation is $1+2i$
Therefore another root is $1-2i$
Sum of the roots=2.
Product=$1+4=5$
The corresponding quadratic factor of the LHS is $x^2-2x+5$
Step 2:
Let $x^2+px+q$ be the other quadratic factor.
Therefore $x^4-4x^3+11x^2-14x+10=(x^2-2x+5)(x^2+px+q)$
Equating the constant
$5q=10$
$\Rightarrow q=2$
Equating the coefficient of $x$
$-14=-4+5p$
$5p=-10$
$p=-2$
The other quadratic factor is $x^2-2x+2$
Step 3:
The other quadratic factor is $x^2-2x+2$
The corresponding roots are obtained by solving $x^2-2x+2=0$
$\Rightarrow x=\large\frac{2\pm\sqrt{4-8}}{2}$
$\Rightarrow 1\pm i$
The roots are $1\pm 2i$,$1\pm i$
edited Jun 11, 2013