How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ ?

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Answer 1 · 2016-08-07T21:08:55

$\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x} = \frac{3}{x} - \frac{5}{x - 2} + \frac{4}{x + 2}$ $(2x^2-18x-12)/(x^3-4x)=3/x-5/(x-2)+4/(x+2)$

Explanation:

$\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$

$= \frac{2 x^{2} - 18 x - 12}{x (x - 2) (x + 2)}$ $=(2x^2-18x-12)/(x(x-2)(x+2))$

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

Use Heaviside's cover-up method to find:

$A = \frac{2 {(0)}^{2} - 18 (0) - 12}{((0) - 2) ((0) + 2)} = \frac{- 12}{- 4} = 3$ $A=(2(0)^2-18(0)-12)/(((0)-2)((0)+2)) = (-12)/(-4) = 3$

$B = \frac{2 {(2)}^{2} - 18 (2) - 12}{(2) ((2) + 2)} = \frac{8 - 36 - 12}{8} = \frac{- 40}{8} = - 5$ $B=(2(2)^2-18(2)-12)/((2)((2)+2)) = (8-36-12)/8 = (-40)/8 = -5$

$C = \frac{2 {(- 2)}^{2} - 18 (- 2) - 12}{(- 2) ((- 2) - 2)} = \frac{8 + 36 - 12}{8} = \frac{32}{8} = 4$ $C=(2(-2)^2-18(-2)-12)/((-2)((-2)-2)) = (8+36-12)/8 = 32/8 = 4$

So:

$\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x} = \frac{3}{x} - \frac{5}{x - 2} + \frac{4}{x + 2}$ $(2x^2-18x-12)/(x^3-4x)=3/x-5/(x-2)+4/(x+2)$

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How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ ?

1 Answer

Explanation:

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How do you write the partial fraction decomposition of the rational expression 2x2−18x−12x3−4x (2x^2-18x-12)/(x^3-4x)?

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Explanation:

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How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ ?