How do you simplify #(x+3)/(x^2-2x-15)#?

1 Answer
Sep 23, 2015

Answer:

The expression simplifies into #1/{x-5}#

Explanation:

You need to find the roots of the denominator: you can either use the classic #{-b \pm \sqrt{b^2-4ac}}/{2a}# formula, our you can use this particular one, which is good only if the coefficient of #x^2# is #1#: in this case, the parabola is of the form #x^2+ax+b#, and you know that #a# equals the sum of the roots, but with the opposite sign, while #b# equals exactly the product of the roots.

So, in this case, we are looking for two roots, which sum up to #2# and which give #-15# when multiplied. This numbers are obviously #5# and #-3#.

Now that you know the roots, you can write down your parabola in a different way: if the roots of #ax^2+bx+c# are #x_0# and #x_1#, then the following holds:

#ax^2+bx+c=(x-x_0)(x-x_1)#.

In your case, since #x_0=5# and #x_1=-3#, you have that

#x^2-2x-15=(x-5)(x+3)#.

You should see that you can cancel the term #(x+3)#: your fraction can be written as

#{x+3}/{x^2-2x-15}=\cancel{x+3}/{(x-5)\cancel((x+3))}=1/{(x-5)}#