Solving for the roots of a polynomial

Question

Suppose we have a polynomial of the form:

$$-x^3+3x^2+9x-27=0$$

Is there an easy way to find the solutions of $x$? I know that they will be factors of $27$, so I begin by factoring $27$ into $1,3,9,27$. I found that $1$ is not a root and $3$ is, so I know that the equation has a factored term $(x-3)$. However, solving the equation using $9$ and $27$ is very tedious and error-prone by hand. Is there a way I can easily find the roots here?

Answer 1

HINT:

We have

$-(-x^3+3x^2+9x-27)$

$=x^3+3^3-3x(x+3)$

$=(x+3)(x^2-3x+3^2)-3x(x+3)$

$=(x+3)\{x^2-3x+3^2-3x\}$

$=(x+3)(x-3)^2$

Answer 2

Since you know that a factor is (x-3), you can use long division or synthetic division to find that the other two roots can be found by the equation f(x)=(-x^2)+9=0. Which of course implies that x^2=9 and x= 3 or -3. So there are in total two roots at 3 and one root at -3.