How do you express $(2x^2-18x-12)/(x^3-4x)$ in partial fractions?

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Answer 1 · 2016-10-30T06:43:37

$(2x^2 - 18x - 12)/(x^3 - 4x) = 3/x + 4/(x + 2) - 5/(x - 2)$

Expand:
$(2x^2 - 18x - 12)/(x^3 - 4x) = A/x + B/(x + 2) + C/(x - 2)$

Multiply both sides by $x(x + 2)(x - 2)$

$2x^2 - 18x - 12 = A(x + 2)(x - 2) + B(x)(x - 2) + C(x)(x + 2)$

Let x = 0 to make B and C disappear:

$-12 = A(2)(-2)$

$A = 3$

Let x = -2 to make A and C disappear:

$2(-2)^2 - 18(-2) - 12 = B(-2)(-2 - 2)$

$32 = B(8)$

$B = 4$

Let x = 2 to make A and B disappear:

$2(2)^2 - 18(2) - 12 = C(2)(2 + 2)$

$-40 = C(8)$

$C = -5$

$(2x^2 - 18x - 12)/(x^3 - 4x) = 3/x + 4/(x + 2) - 5/(x - 2)$

How do you express (2x^2-18x-12)/(x^3-4x)(2x^2-18x-12)/(x^3-4x) in partial fractions?