How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{(x^{2} - x - 6) (x + 2)}$ $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{(x^{2} - x - 6) (x + 2)}$ $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))$ ?

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Answer 1 · 2017-08-01T04:44:02

$\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{(x^{2} - x - 6) (x + 2)} = 2 + \frac{1}{x - 3} - \frac{3}{x + 2} + \frac{2}{{(x + 2)}^{2}}$ $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))=2+1/(x-3)-3/(x+2)+2/(x+2)^2$

Explanation:

Before we express in partial fractions, ensure that the degree of numerator is less than that of denominator.

$\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{(x^{2} - x - 6) (x + 2)}$ $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))$

= $\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{x^{3} + x^{2} - 8 x - 12}$ $(2x^3+2x^2-9x-20)/(x^3+x^2-8x-12)$

= $2 + \frac{7 x + 4}{x^{3} + x^{2} - 8 x - 12}$ $2+(7x+4)/(x^3+x^2-8x-12)$

= $2 + \frac{7 x + 4}{(x - 3) (x + 2) (x + 2)}$ $2+(7x+4)/((x-3)(x+2)(x+2))$

= $2 + \frac{7 x + 4}{(x - 3) {(x + 2)}^{2}}$ $2+(7x+4)/((x-3)(x+2)^2)$

Now let $\frac{7 x + 4}{(x - 3) {(x + 2)}^{2}} \equiv \frac{A}{x - 3} + \frac{B}{x + 2} + \frac{C}{{(x + 2)}^{2}}$ $(7x+4)/((x-3)(x+2)^2)-=A/(x-3)+B/(x+2)+C/(x+2)^2$

or $\frac{7 x + 4}{(x - 3) {(x + 2)}^{2}} = \frac{A {(x + 2)}^{2} + B (x - 3) (x + 2) + C (x - 3)}{(x - 3) {(x + 2)}^{2}}$ $(7x+4)/((x-3)(x+2)^2)=(A(x+2)^2+B(x-3)(x+2)+C(x-3))/((x-3)(x+2)^2)$

if $x = 3$ $x=3$ , then $25 A = 25$ $25A=25$ i.e. $A = 1$ $A=1$

if $x = - 2$ $x=-2$ , then $- 5 C = - 10$ $-5C=-10$ i.e. $C = 2$ $C=2$

and comparing coefficients of $x^{2}$ $x^2$ on both sides

$A + B + C = 0$ $A+B+C=0$ i.e. $B = - 3$ $B=-3$

Hence

$\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{(x^{2} - x - 6) (x + 2)} = 2 + \frac{1}{x - 3} - \frac{3}{x + 2} + \frac{2}{{(x + 2)}^{2}}$ $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))=2+1/(x-3)-3/(x+2)+2/(x+2)^2$

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How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{(x^{2} - x - 6) (x + 2)}$ $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))$ ?

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How do you write the partial fraction decomposition of the rational expression 2x3+2x2−9x−20(x2−x−6)(x+2)(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))?

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How do you write the partial fraction decomposition of the rational expression $\frac{2 x^{3} + 2 x^{2} - 9 x - 20}{(x^{2} - x - 6) (x + 2)}$ $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))$ ?