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

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Answer 1 · 2015-05-12T02:29:51

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

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

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

$x + 1 = 0$ $x+1=0$

Giving:
$x_{1} = 3$ $x_1=3$
$x_{2} = - 3$ $x_2=-3$
$x_{3} = - 1$ $x_3=-1$

How do you find all zeros of y=x3+x2−9x−9y=x^3+x^2-9x-9?