# How do you find all zeros of y=x^3+x^2-9x-9?

##### 1 Answer
May 12, 2015

Set $y = 0$ and solve for $x$:
${x}^{3} + {x}^{2} - 9 x - 9 = 0$

Collecting $x$ from the first and third term:
$x \left({x}^{2} - 9\right) + {x}^{2} - 9 = 0$
$\left({x}^{2} - 9\right) \left(x + 1\right) = 0$
So:

${x}^{2} - 9 = 0$
$x = \pm \sqrt{9} = \pm 3$
and

$x + 1 = 0$

Giving:
${x}_{1} = 3$
${x}_{2} = - 3$
${x}_{3} = - 1$