How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 10 x - 5}{{(x + 1)}^{2} (x - 2)}$ $(3x^2 + 10x -5)/ ( (x+1)^2(x-2) )$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 10 x - 5}{{(x + 1)}^{2} (x - 2)}$ $(3x^2 + 10x -5)/ ( (x+1)^2(x-2) )$ ?

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Answer 1 · 2016-12-30T09:43:40

The answer is $= \frac{4}{{(x + 1)}^{2}} + \frac{3}{x - 2}$ $=4/(x+1)^2+3/(x-2)$

Explanation:

Let's do the decomposition into partial fractions

$\frac{3 x^{2} + 10 x - 5}{{(x + 1)}^{2} (x - 2)} = \frac{A}{{(x + 1)}^{2}} + \frac{B}{x + 1} + \frac{C}{x - 2}$ $(3x^2+10x-5)/((x+1)^2(x-2))=A/(x+1)^2+B/(x+1)+C/(x-2)$

$= \frac{A (x - 2) + B (x + 1) (x - 2) + C {(x + 1)}^{2}}{{(x + 1)}^{2} (x - 2)}$ $=(A(x-2)+B(x+1)(x-2)+C(x+1)^2)/((x+1)^2(x-2))$

Therefore,

$3 x^{2} + 10 x - 5 = A (x - 2) + B (x + 1) (x - 2) + C {(x + 1)}^{2}$ $3x^2+10x-5=A(x-2)+B(x+1)(x-2)+C(x+1)^2$

Let $x = - 1$ $x=-1$ , $\Rightarrow$ $=>$ , $- 12 = - 3 A$ $-12=-3A$ , $\Rightarrow$ $=>$ , $A = 4$ $A=4$

Ler $x = 2$ $x=2$ , $\Rightarrow$ $=>$ , $27 = 9 C$ $27=9C$ , $\Rightarrow$ $=>$ , $C = 3$ $C=3$

Coefficients of $x^{2}$ $x^2$

$3 = B + C$ $3=B+C$ , $\Rightarrow$ $=>$ , $B = 3 - C = 0$ $B=3-C=0$

So,

$\frac{3 x^{2} + 10 x - 5}{{(x + 1)}^{2} (x - 2)} = \frac{4}{{(x + 1)}^{2}} + \frac{0}{x + 1} + \frac{3}{x - 2}$ $(3x^2+10x-5)/((x+1)^2(x-2))=4/(x+1)^2+0/(x+1)+3/(x-2)$

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How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 10 x - 5}{{(x + 1)}^{2} (x - 2)}$ $(3x^2 + 10x -5)/ ( (x+1)^2(x-2) )$ ?

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Explanation:

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How do you write the partial fraction decomposition of the rational expression (3x^2 + 10x -5)/ ( (x+1)^2(x-2) )3x2+10x−5(x+1)2(x−2)(3x^2 + 10x -5)/ ( (x+1)^2(x-2) )?

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How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 10 x - 5}{{(x + 1)}^{2} (x - 2)}$ $(3x^2 + 10x -5)/ ( (x+1)^2(x-2) )$ ?