20 PTS! Help!
How do you solve the following:

Answer: [tex]3x^2-x+2[/tex]
Explanation:
The area of a triangle A can be calculated as A=(height)x(base)/2
We know the height is 6x.
We know [tex]A=9x^3-3x^2+6x[/tex]
So the equation becomes (using |AC| as the base length):
[tex]9x^3-3x^2+6x = \frac{6x \cdot |AC|}{2}\\\frac{9x^3-3x^2+6x}{3x}=|AC|\\\frac{3x^2-x+2}{1}=|AC|\\|AC|=3x^2-x+2[/tex]
The base is [tex]3x^2-x+2[/tex]