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

1 Answer
Mar 5, 2017

Answer:

Rewrite in single-line form. Factor. Cancel common factors. Re-distribute if required.
The reduced expression is #(p+3)/((p-2)(p+1))#.

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 #(p+a)(p+b)#, you'll get #p^2+(a"+"b)p+ab.# Factoring means working backwards from #p^2+cp+d# by trying to find the numbers #a# and #b# such that #a+b=c# and #ab=d#.

Starting with #p^2+p-6#, this can be factored to #(p+3)(p-2)# (since #3xx"-"2="-"6# and #3+("-"2)=1#).

Similarly, #p^2+6p+9=(p+3)(p+3)=(p+3)^2#,

and #p^2+4p+3=(p+3)(p+1)#.

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 #p+3# factors—three on the top, and two on the bottom. Thus, we can cancel off two pairs, like this:

#=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 #p+3# as a factor in its denominator. Thus, if #p+3# were to equal 0, we'd get division by zero, which is undefined. The simplified expression does not have #p+3# as a factor in its denominator. In order to have the simplified expression reflect the original one 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#