How do you simplify #(x^2+3x+2)/(x^2-1)#?

2 Answers

Answer:

#(x²+3x+2)/(x²-1)=(x+2)/(x-1)#

Explanation:

#(x²+3x+2)/(x²-1)#
#=(cancel((x+1)) (x+2))/((x-1)cancel((x+1)))#
#=(x+2)/(x-1)#
\0/ here's our answer!

May 15, 2018

First we should split the middle term

By the video... you should have understood that we need to make #3x# in two number that one is divisible by 2 and the other is divisible itself

#(x^2+x+2x+2)/(x^2-1)#

You should always remember that
#1=1^2#
It'll always help you

Factorize
#(x(x+1)+2(x+1))/(x^2-1^2)#

It is a law of exponents
#a^2-b^2=(a+b)(a-b)#

#((x+2)cancel((x+1)))/((x-1)cancel((x+1)))#

You are left with
#(x+2)/(x-1)#