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How do you simplify the following expression:

$$\frac{x^3-27}{x^2+x-6}$$

Thanks for the help. Haven't seen this stuff since high school and I'm trying to help my younger sister out.

OR.
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JustAnotherCoder
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3 Answers3

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There are two types of transformations that normally look like they simplify the rational expression.

  1. One is turning an improper fraction (the degree of the numerator is larger than or equal to the one of the denominator) into a sum of a polynomial plus a proper fraction (the degree of the numerator is strictly smaller than the one of the denominator).
  2. The other is dividing numerator and denominator by their greatest common divisor.

Both are subproducts of Euclid's algorithm, which main tool is polynomial long division.

There are other transformations of rational functions like partial fraction decomposition, but visually it may not always look like simplifying or depending on the school grade they may not be supposed to know it.

Let me explain how 1 and 2 go:

To do $1$ we do polynomial long division to write $x^3-27=(x^2+x-6)q(x)+r(x)$, where $q$ and $r$ are two polynomials, and the degree of $r$ is $<3$. This gives you the simplification $$\frac{x^3-27}{x^2+x-6}=q(x)+\frac{r(x)}{x^2+x-6}.$$

I got lazy. It is written here. But first master doing polynomial long division.

For $2$, the full Euclid's algorithm, of which the step $1$ is the first computation, gives you a polynomial $d(x)$ of maximum degree that divides both $r(x)$ and $x^2+x-6$. Then the final simplification could be to write $$\frac{x^3-27}{x^2+x-6}=q(x)+\frac{r(x)/d(x)}{(x^2+x-6)/d(x)}.$$

People are asking you about the signs because in you problem $d(x)=1$ and this turns step $2$ in trivial, i.e. there is not going to be more simplification beyond step $1$.

From the long division you should get $q(x)=x-1$ and $r(x)=7x-33$. This gives $$\frac{x^3-27}{x^2+x-6}=x-1+\frac{7x-33}{x^2+x-6}.$$

Finally simplification is a relative notion. An expression is simpler depending on what you need it for afterwards. Therefore other transformations could also be considered simplifications.

OR.
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1

The general idea would be to factor, though there aren't any common factors here:

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

However, if that + in the denominator is the wrong sign, then this can be reduced, assuming $x\neq3$:

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

Or one could flip the sign in the numerator for the 27, assume $x\neq-3$ and then:

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

JB King
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    @JustAnotherCoder Two useful factorizations which aren't taught nearly often enough are the sum and difference of two cubes: $ \ x^3 + a^3 = (x + a) \cdot (x^2 - ax + a^2) \ $ and $ \ x^3 - a^3 = (x - a) \cdot (x^2 + ax + a^2) \ $ . – colormegone Jul 31 '13 at 20:59
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If you plug $3$ into a polynomial and get $0$, then $x-3$ is one of the factors.

You have $x^3-27$. Plug in $3$ and get $3^3-27=0$.

So $x^3-27 = (x-3)(\cdots\cdots\cdots)$. Then you can fill in the blank by using long division to divide $x^3-27$ by $x-3$.

Similarly $x^2-x-6$ becomes $3^2-3-6=0$, so you can factor: $x^2-x-6=(x-3)(\cdots\cdots\cdots)$, and again you need to find the other factor.

If you get $\dfrac{(x-3)(\cdots\cdots\cdots)}{(x-3)(\cdots\cdots\cdots)}$, then you can cancel $x-3$.

Michael Hardy
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