Question

How do you simplify `\frac { x ^ { 2} + 9x + 20} { x ^ { 2} - 2x - 24}`?

Answer 1

Factor any possible terms. Cross out if possible. The result is `(x+5)/(x-6)`

Explanation:

First, let's factor the numerator and denominator.

`(x^2+9x+20)/(x^2-2x-24)`

It's a simple trinomial.

`((x+4)(x+5))/((x+4)(x-6)`

and we cross out `(x+4)`

`(cancel((x+4))(x+5))/(cancel((x+4))(x-6)`

and result is `(x+5)/(x-6)`

Answer 2

`(x^2+9x+20)/(x^2-2x-24) = 1+11/(x-6)`

with exclusion `x != -4`

Explanation:

Given:

`(x^2+9x+20)/(x^2-2x-24)`

If the numerator of a rational expression is of degree greater than or equal to the denominator, then we can simplify a little before factoring. In the given example we could almost as easily factor both first, but the following approach is helpful to break down more complicated examples.

In our example we find:

`(x^2+9x+20)/(x^2-2x-24) = ((x^2-2x-24)+11x+44)/(x^2-2x-24)`

`color(white)((x^2+9x+20)/(x^2-2x-24)) = (x^2-2x-24)/(x^2-2x-24)+(11x+44)/(x^2-2x-24)`

`color(white)((x^2+9x+20)/(x^2-2x-24)) = 1+(11(x+4))/(x^2-2x-24)`

Now that we have simplified the numerator, we can check whether `(x+4)` is a factor of the denominator, finding:

`color(white)((x^2+9x+20)/(x^2-2x-24)) = 1+(11color(red)(cancel(color(black)((x+4)))))/(color(red)(cancel(color(black)((x+4))))(x-6))`

`color(white)((x^2+9x+20)/(x^2-2x-24)) = 1+11/(x-6)`

with exclusion `x != -4`

I mention the exclusion since the original rational expression is undefined when `x=-4`, while the simplified expression is defined.