The roots of $x^4+4x^3+5x^2+2x+2=0$ one root being $-1+i$ are

Question

The roots of $x^4+4x^3+5x^2+2x+2=0$, one root being $-1+i$ are what?

please solve this problem, i need the process of solution

Answer 1

Hint: If $-1+i$ is a root, then $-1-i$ is also a root. Now do polynomial long division by $(x+1-i)(x+1+i)$, then solve the remaining quadratic.

However, it seems like $-1+i$ is actually not a root of your polynomial. Did you make an error typing it up?

Answer 2

When $x = (-1 + i)$, $x^4+4x^3+5x^2+2x+2 = 4$. Since $x^4 = -4$, I will assume that the correct polynomial in question is $$2x^4+4x^3+5x^2+2x+2.$$ Since one root is $-1 + i$, it follows that its conjugate, $-1 - i$, is also a root, and so $(x+1-i)(x+1+i)$ is a factor of the polynomial. Note that $(x+1-i)(x+1+i)$ is the difference of $(x+1)^2$ and $i^2$; $(x+1-i)(x+1+i) = x^2+2x+2$ . Using long division we can see that $$\frac {2x^4+4x^3+5x^2+2x+2}{x^2+2x+2} = 2x^2 + 1$$ The solutions of $x^2 = -\frac 12$ are $\sqrt{-1/2} = \pm \frac {i}{\sqrt{2}} = \pm \frac {\sqrt{2}i}{2}$. Thus the roots of $2x^4+4x^3+5x^2+2x+2 = 0$ are $$x = -1+i, -1-i, \frac {\sqrt{2}i}{2}, -\frac {\sqrt{2}i}{2}$$