How do you express $(5x-1)/(x^2-x-2)$ in partial fractions?

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Answer 1 · 2016-11-29T21:46:55

Factorize the denominator, then calculate the numerators of partial fractions.

First you find the zeroes of the denominator to factorize it.

$x^2-x-2 =0$

$x= frac (1+-sqrt(1+8)) 2 = (1+-3)/2$

$x^2-x-2 = (x-2)(x+1)$

Now pose:

$frac (5x-1) ((x-2)(x+1)) = frac A (x-2) + frac B (x+1)$

$frac (5x-1) ((x-2)(x+1)) = frac ( A(x+1)+B(x-2)) ((x-2)(x+1))$

$frac (5x-1) ((x-2)(x+1)) = frac ( Ax+A+Bx-2B) ((x-2)(x+1))$

Equating the coefficient of the same order in the numerators:

$A+B=5$
$A-2B=-1$

and solving the system:

$A=3$
$B=2$

thus:

$frac (5x-1) ((x-2)(x+1)) = frac 3 (x-2) + frac 2 (x+1)$

How do you express (5x-1)/(x^2-x-2)(5x-1)/(x^2-x-2) in partial fractions?