What is easiest method to get solutions for $(1-x)(-x^3+x^2+9x-9)=0$?

Question

Okay, first $x_1=1$, but what is easiest method to get other 3 solutions?

Answer 1

You can factor the terms in the second parenthesis, i.e. $$-x^3 + x^2 +9x- 9 = -x^2(x-1) +9(x-1) = (-x^2 +9) (x-1) = -(x^2-9)(x-1).$$

Answer 2

The possible integer roots of $-x^3+x^2+9x-9$ are the divisors of $-9$, so $\{-9,-3,-1,1,3,9\}$, $3$ is such a root so $-x^3+x^2+9x-9=(x-3)(-x^2-2x+3)$. Finally $(1-x)(-x^3+x^2+9x-9)=-(1-x)(x-3)(x^2+2x-3)$. You just have to solve a quadratic equation. Its root are $1$ and $-3$.

Finally $(1-x)(-x^3+x^2+9x-9)=-(1-x)(x-3)(x-1)(x+3)$.