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

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

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Answer 1 · 2016-07-29T20:03:43

$\frac{x}{x^{3} - x^{2} - 6 x} = \frac{1}{5 (x - 3)} - \frac{1}{5 (x + 2)}$ $x/(x^3-x^2-6x)=1/(5(x-3))-1/(5(x+2))$

with exclusion $x \neq 0$ $x != 0$

Explanation:

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

$\frac{x}{x^{3} - x^{2} - 6 x}$ $x/(x^3-x^2-6x)$

$= \frac{1}{x^{2} - x - 6}$ $=1/(x^2-x-6)$

$= \frac{1}{(x - 3) (x + 2)}$ $=1/((x-3)(x+2))$

$= \frac{A}{x - 3} + \frac{B}{x + 2}$ $=A/(x-3)+B/(x+2)$

with exclusion $x \neq 0$ $x != 0$

Use Heaviside's cover-up method to find:

$A = \frac{1}{(3) + 2} = \frac{1}{5}$ $A = 1/((3)+2) = 1/5$

$B = \frac{1}{(- 2) - 3} = - \frac{1}{5}$ $B = 1/((-2)-3) = -1/5$

So:

$\frac{x}{x^{3} - x^{2} - 6 x} = \frac{1}{5 (x - 3)} - \frac{1}{5 (x + 2)}$ $x/(x^3-x^2-6x)=1/(5(x-3))-1/(5(x+2))$

with exclusion $x \neq 0$ $x != 0$

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

1 Answer

Explanation:

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