(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 .
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Category: computerinformation |
Author: Selma Yafa

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