How do you simplify #(x^2+4x-5)/(x^2-1)#?

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Answer 1 · 2017-01-24T14:26:27

Factor the numerator and the denominator, and see if you can cancel any of the factors!

#((x+5)(x-1))/((x+1)(x-1))# reduces by canceling out the repeated (x-1) that appears in both top and bottom of the expression.

Your final answer is: #(x+5)/(x+1)#.

Answer 2 · 2017-01-24T14:28:46

The answer is #=(x+5)/(x+1)#

We need

#a^2-b^2=(a+b)(a-b)#

Let's factorise the numerator and denominator

#x^2+4x-5=(x-1)(x+5)#

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

Therefore,

#(x^2+4x-5)/(x^2-1)=(cancel(x-1)(x+5))/((x+1)cancel(x-1))#

#=(x+5)/(x+1)#