Simplifying $\frac{2x^3-9x^2+27}{3x^3-81x+162} $ Simplify $$\frac{2x^3-9x^2+27}{3x^3-81x+162} $$
All I can see is thus far is the factor 3 in the denominator can be taken out. Then I am stuck because I don't recognize any of the usual patterns in simplification.
The answer is $\;\dfrac{2x+3}{3(x+6)}\;,\;$ so clearly I am missing something here. Thanks.

By the rational roots theorem, the rational roots of the numerator are $-\frac32$ and $3$ and, if you divide the numerator by $\left(x+\frac32\right)(x-3)$, you will get $2x-6$. Therefore, the numerator is equal to $(2x+3)(x-3)^2$. On the other hand, the rational roots of the denominator are $-6$ and $3$; in fact, the denominator is equal to $3(x+6)(x-3)^2$. Therefore, the quotient is indeed $\frac{2x+3}{3(x+6)}$.
