Q.Solve the equation $x^4-4x^2+8x+35=0$, if one of its roots is $2+\sqrt3\,i$.

Use the conjugate-root theorem to get a quadratic factor, divide it out of the quartic, then solve the remaining quadratic.

1. Conjugate root. The polynomial $x^4-4x^2+8x+35=0$ has real coefficients, so if $2+\sqrt3\,i$ is a root, its conjugate $2-\sqrt3\,i$ is also a root.

2. Form the quadratic factor from this conjugate pair.

Sum $=(2+\sqrt3 i)+(2-\sqrt3 i)=4$; Product $=(2+\sqrt3 i)(2-\sqrt3 i)=4+3=7$.

$x^2-(\text{sum})x+\text{product}=x^2-4x+7$

3. Divide the quartic by this factor.

$x^4+0x^3-4x^2+8x+35 \div (x^2-4x+7)$

• $x^4\div x^2=x^2$; $x^2(x^2-4x+7)=x^4-4x^3+7x^2$; remainder: $4x^3-11x^2+8x+35$
• $4x^3\div x^2=4x$; $4x(x^2-4x+7)=4x^3-16x^2+28x$; remainder: $5x^2-20x+35$ …