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

Since the coefficients are real, $1-i$ (the conjugate) is also a root; divide out the quadratic factor these two roots give, then solve the remaining quadratic.

Because all coefficients of $x^4+2x^3-5x^2+6x+2=0$ are real, complex roots occur in conjugate pairs, so $1-i$ is also a root alongside the given $1+i$. These two roots give the quadratic factor

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

Divide the quartic by this factor:

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

(Check: expanding the right side reproduces $x^4+2x^3-5x^2+6x+2$ exactly.)

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