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,

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

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

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