Question

HELP HELP HELP HELP HELP HELP (STEP BY STEP)

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

$\dfrac{2x^3 - 3x^2 - x + 4}{x^2 - 1} = \dfrac{(2x-3)(x-1)(x+1) + x+1}{(x-1)(x+1)} \\\\ ~~~~~~~~ = \boxed{2x-3 + \dfrac1{x-1}}$