Question

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Answer 1

With some simple rearrangement, we can rewrite the numerator as

`2x^3 - 3x^2 - x + 4 = 2(x^3 - x) - 3x^2 + x + 4 \\\\ ~~~~~~~~ = 2x(x^2-1) - 3(x^2 - 1) + x + 1 \\\\ ~~~~~~~~ = (2x-3)(x^2-1) + x+1`

Then factorizing the difference of squares,
`x^2-1=(x-1)(x+1)`, we end up with

`(2x^3 - 3x^2 - x + 4)/(x^2 - 1) = ((2x-3)(x-1)(x+1) + x+1)/((x-1)(x+1)) \\\\ ~~~~~~~~ = \boxed{2x-3 + \frac1{x-1}}`