Q.Solve the equation $x^4 + 2x^3 - 5x^2 + 6x + 2 = 0$, given that $1 + i$ is one of its roots.

The four roots are $1\pm i$ and $-2\pm\sqrt3$.

Since the coefficients are real and $1+i$ is a root, its conjugate $1-i$ is also a root.

The corresponding quadratic factor is

$\big(x-(1+i)\big)\big(x-(1-i)\big)=x^2-2x+2.$

Divide $x^4+2x^3-5x^2+6x+2$ by $x^2-2x+2$:

$x^4+2x^3-5x^2+6x+2=(x^2-2x+2)(x^2+4x+1).$

…