How do you express $\frac{x - 1}{x^{3} + x^{2}}$ $(x-1)/(x^3 +x^2)$ in partial fractions?

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How do you express $\frac{x - 1}{x^{3} + x^{2}}$ $(x-1)/(x^3 +x^2)$ in partial fractions?

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Answer 1 · 2016-10-29T20:16:18

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

Explanation:

Some factorisation to start with
$\frac{x - 1}{x^{3} + x^{2}} = \frac{x - 1}{(x^{2}) (x + 1)}$ $(x-1)/(x^3+x^2)=(x-1)/((x^2)(x+1))$
$= \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x + 1}$ $=A/x+B/x^2+C/(x+1)$
$= \frac{A x (x + 1) + B (x + 1) + C x^{2}}{(x^{2}) (x + 1)}$ $=(Ax(x+1)+B(x+1)+Cx^2)/((x^2)(x+1))$

So now we can solve for $A, B, and C$ $A, B,and C$
$x - 1 = A x (x + 1) + B (x + 1) + C x^{2}$ $x-1=Ax(x+1)+B(x+1)+Cx^2$
let x=-1, $- 2 = C$ $-2=C$ $\Rightarrow$ $=>$ $C = - 2$ $C=-2$
Coefficients of $x^{2}$ $x^2$ , $0 = A + C$ $0=A+C$ $\Rightarrow$ $=>$ $A = 2$ $A=2$
Coefficients of $x$ $x$ , $1 = A + B$ $1=A+B$ $\Rightarrow$ $=>$ $B = - 1$ $B=-1$

And finally, we have
$\frac{x - 1}{x^{3} + x^{2}} = \frac{2}{x} - \frac{1}{x^{2}} - \frac{2}{x + 1}$ $(x-1)/(x^3+x^2)=2/x-1/x^2-2/(x+1)$

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How do you express $\frac{x - 1}{x^{3} + x^{2}}$ $(x-1)/(x^3 +x^2)$ in partial fractions?

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Explanation:

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How do you express $\frac{x - 1}{x^{3} + x^{2}}$ $(x-1)/(x^3 +x^2)$ in partial fractions?