How do you write the partial fraction decomposition of the rational expression $x/(x^3-x^2-6x)$ ?

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Answer 1 · 2016-07-29T20:03:43

$x/(x^3-x^2-6x)=1/(5(x-3))-1/(5(x+2))$

with exclusion $x != 0$

Both the numerator and the denominator are divisible by $x$ , so we can simplify (noting $x=0$ as an excluded value):

$x/(x^3-x^2-6x)$

$=1/(x^2-x-6)$

$=1/((x-3)(x+2))$

$=A/(x-3)+B/(x+2)$

with exclusion $x != 0$

Use Heaviside's cover-up method to find:

$A = 1/((3)+2) = 1/5$

$B = 1/((-2)-3) = -1/5$

So:

$x/(x^3-x^2-6x)=1/(5(x-3))-1/(5(x+2))$

with exclusion $x != 0$

How do you write the partial fraction decomposition of the rational expression x/(x^3-x^2-6x) x/(x^3-x^2-6x)?