(x)(x2−1)
Now we can see that the (x2−1) can be further factored.
x+1
(x)(x−1)(x+1)
We now have two expressions of (x+1), one on the numerator and one on the denominator, which means we can cancel them out and simply put 1 in the numerator.
1
x(x−1)
And once we distribute the x back in the denominator, we will have:
1
x2−x
Our final answer is J,
1
x2−x
.
