# 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.

• Have you tried finding the roots of the polynomials? $x=3$ is a root for both polynomials Commented Jan 12, 2023 at 19:32
• $2x^3-9x^2+27=(2x+3)(x-3)^2$ and $3x^3-81x+162=3(x+6)(x-3)^2.\;$ Commented Jan 12, 2023 at 19:34
• Use polynomial Euclidean algorithm to find and cancel the gcd of the two polynomials.
– lhf
Commented Jan 12, 2023 at 19:50

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 $$\frac{2x+3}{3(x+6)}$$.

• Can't we find the roots via the cubic formula $?$ Commented Jan 12, 2023 at 19:41
• @MathStackexchangeIsVeryBad Sure. And we can also ill a mosquito with a cannon. Commented Jan 12, 2023 at 19:44
• +1...i really liked the reference Commented Jan 12, 2023 at 19:46

I know this is an old one but I would give a third option in addition to the previous solutions, just to show another possible way of reasoning.

$$\frac{2x^3-9x^2+27}{3x^2-81x+162} = \frac{1}{3}\frac{2x^3-9x^2+27}{x^3-27x+54} = \frac{1}{3}\frac{(x^3-27x+54) + (x^3-9x^2+27x-27)}{x^2-27x+54} = \frac{1}{3}\left(1+\frac{x^3-9x^2+27x-27}{x^3-27x+54}\right)$$

Now we can factorize observing that $$x^3-9x^2+27x-27 = (x-3)^3$$ and using polynomial division by $$x-3$$ on the denominator and then factorizing we get $$x^3-27x+54 = (x-3)(x^2+3x+18) = (x-3)^2(x+6)$$. Plugging these where we left we finally get

$$\frac{1}{3}\left(1+\frac{x^3-9x^2+27x-27}{x^3-27x+54}\right) = \frac{1}{3}\left(1+\frac{(x-3)^3}{(x-3)^2(x+6)}\right) = \frac{1}{3}\left(1+\frac{(x-3)^3}{(x-3)^2(x+6)}\right) = \frac{1}{3}\left(1+\frac{x-3}{x+6}\right) = \frac{2x+3}{3(x+6)}$$

• How did you know to break up $2x^3-9x^2+27$ into $(x^3-27x+54)+(x^3-9x^2+27x-27)$? Is it because you wanted to match the denominator and cancel it? But then this only works in this case because $(x^3-9x^2+27x-27)$ happens to be $(x-3)^3$. I guess what I'm saying is obviously this requires some special intuition or knowledge which I do not possess. So when I look at these kinds of problems, I would never be able to perform this kind of creativity and flexibility. Can you explain to me how you did it and what cues/knowledge/intuition you relied on? Thank you. Commented Aug 4, 2023 at 5:15
• Sure, without external knowledge, trying to find an easier rewriting may become a matter of trial and error if we don't want to use more sophisticated techniques. In this case trying to split the numerator into something that matches the denominator plus something is one of those common math tricks. The intuition on why this may have worked for me was that the two polynomials had the same degree (hence we cannot simply devide one by the other) and that polynomials with a factor $1$ for the higher degree term are usually easier to deal with. Commented Aug 4, 2023 at 7:24

The polynomial Euclidean algorithm gives $$\gcd(2x^3-9x^2+27,3x^3-81x+162)=x^2 - 6 x + 9=(x - 3)^2$$ Now cancel this factor in the numerator and in the denominator to reduce the fraction to the given answer.