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

1 Answer
Feb 5, 2015

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

First rewrite -8x^2 as -9x^2+x^2.

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

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

This gives us x^3-9x^2+18x+x^2-9x+18

Now we are ready to factor.

x(x^2-9x+18)+(x-6)(x-3)=0
x(x-6)(x-3)+(x-6)(x-3)=0
(x-6)(x-3)(x+1)=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