How do you factor complete x^3-8x^2+9x+18=0?

Feb 5, 2015

We need to rewrite the polynomial equation so that we can properly factor it.

First rewrite $- 8 {x}^{2}$ as $- 9 {x}^{2} + {x}^{2}$.

This gives us ${x}^{3} - 9 {x}^{2} + {x}^{2} + 9 x + 18 = 0$

We now need to rewrite $9 x$ as $18 x - 9 x$

This gives us ${x}^{3} - 9 {x}^{2} + 18 x + {x}^{2} - 9 x + 18$

Now we are ready to factor.

$x \left({x}^{2} - 9 x + 18\right) + \left(x - 6\right) \left(x - 3\right) = 0$
$x \left(x - 6\right) \left(x - 3\right) + \left(x - 6\right) \left(x - 3\right) = 0$
$\left(x - 6\right) \left(x - 3\right) \left(x + 1\right) = 0$

Once the equation is factored we can set each factor equal to 0 to find the solutions, x-intercepts, to the polynomial equation.

$x - 6 = 0$
$x = 6$

$x - 3 = 0$
$x = 3$

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