Find all the roots of equation whether real or complex

Question

$$f(x)=(x^4 +7x^3 +22x^2 + 31x + 9)^8$$ Find all the roots of the equation, given $-2+\sqrt5i$ is a root to this equation $f(x)$.

Answer 1

Notice that, for a polynomial with real coefficients, if $z\in\mathbb{C}$ is a root, then $\overline{z}$ is also a root. At your case, $z = -2+i\sqrt{5}$ is known to be a root. Then divide the fourth degree polynomial by $(x-z)(x-\overline{z})$, from whence you obtain a second degree equation. Can you take it from here?

Answer 2

First, since $0^8= 0$ and, conversely, if $x^8= 0$, x= 0, any x is a root of $(x^4+ 7x^3+ 22x^2+ 31x+ 9)^2= 0$ is a root of $x^4+ 7x^3+ 22x^2+ 31x+ 9= 0$. Further, since all coefficients are real numbers, given that $-2+ i\sqrt{5}$, $-2- i\sqrt{5}$ is also. So $(x+ 2- i\sqrt{5})(x+ 2+ i\sqrt{5})= x^2+ 4x+ 4+ 5= x^2+ 4x+ 9$ is a factor. Dividing $x^4+ 7x^3+ 22x^2+ 31x+ 9$ by $x^2+ 4x+ 4+ 5= x^2+ 4x+ 9$ gives you a quadratic equation you can solve for the other two roots.