# How do you simplify #((p+3)/(p^2+p-6))/((p^2+4p+3)/(p^2+6p+9))#?

##### 1 Answer

#### Answer:

Rewrite in single-line form. Factor. Cancel common factors. Re-distribute if required.

The reduced expression is

#### Explanation:

First, let's get rid of that nasty embedded fraction look:

#(" "(p+3)/(p^2+p-6)" ")/((p^2+4p+3)/(p^2+6p+9))#

#=(p+3)/(p^2+p-6)-:(p^2+4p+3)/(p^2+6p+9)#

Then, we can use the rule that *division is just multiplication by a reciprocal* to get:

#=(p+3)/(p^2+p-6) xx (p^2+6p+9)/(p^2+4p+3)#

Alright, now it's time to start factoring!

We're going to look at each numerator (or denominator) in turn, and reduce each to its factored form. Remember how, if you were to distribute something like *backwards* from

Starting with

Similarly,

and

Making these substitutions, our original expression becomes

#=(p+3)/((p+3)(p-2)) xx ((p+3)(p+3))/((p+3)(p+1))#

At this point, because we've rewritten the expression into its multiplicative factors, identical factors can be cancelled off in pairs, as long as they're on opposite sides of the fraction lines.

There are a lot of

#=color(blue)cancel(color(black)(p+3))/(color(blue)(cancel(color(black)((p+3))))(p-2)) xx (color(red)(cancel(color(black)((p+3))))(p+3))/(color(red)(cancel(color(black)((p+3))))(p+1))#

#=1/(p-2) xx (p+3)/(p+1)#

#=(p+3)/((p-2)(p+1))#

Often, this is as far as we'll go, since factors are more useful than polynomials (as we've just seen). But, if you wanted to distribute the final denominator, you could, to get:

#=(p+3)/(p^2-p-2)# .

## Bonus:

The original expression has *does not* have *exactly*, we must state this restriction alongside our simplified expression, as follows:

#(" "(p+3)/(p^2+p-6)" ")/((p^2+4p+3)/(p^2+6p+9))=(p+3)/((p-2)(p+1))," " p+3!=0#