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

Since the polynomial has real coefficients, $1-i$ is also a root; divide out the quadratic factor these two roots produce, then solve the remaining quadratic.

Given $1+i$ is a root of $x^4+2x^3-5x^2+6x+2=0$, and the coefficients are all real, the conjugate $1-i$ must also be a root (complex roots of a real polynomial occur in conjugate pairs).

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 $x^2-2x+2$:

$x^4+2x^3-5x^2+6x+2 \div (x^2-2x+2):$

• $x^4\div x^2=x^2$; subtract $x^2(x^2-2x+2)=x^4-2x^3+2x^2$: remainder $4x^3-7x^2+6x+2$.
• $4x^3\div x^2=4x$; subtract $4x(x^2-2x+2)=4x^3-8x^2+8x$: remainder $x^2-2x+2$. …