How do you express $(x^2)/(x+1)^3$ in partial fractions?

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Answer 1 · 2016-02-18T07:52:27

$x^2/(x+1)^3=1/(x+1)^3-2/(x+1)^2+1/(x+1)$

Set up the equation first by assigning letter variables A, B, C.

$x^2/(x+1)^3=A/(x+1)^3+B/(x+1)^2+C/(x+1)$

$x^2/(x+1)^3=A/(x+1)^3+(B(x+1))/(x+1)^3+(C(x+1)^2)/(x+1)^3$

$x^2/(x+1)^3=(A+B(x+1)+C(x+1)^2)/(x+1)^3$

$x^2/(x+1)^3=(A+Bx+B+Cx^2+2Cx+C)/(x+1)^3$

$(x^2+0*x+0*x^0)/(x+1)^3=(Cx^2+Bx+2Cx+(A+B+C)x^0)/(x+1)^3$

Set up the equations to solve for the values of A, B, C by equating the coefficients of each term

$Cx^2=1*x^2$

$Bx+2Cx=0*x$

$(A+B+C)x^0=0*x^0$

Therefore, after simplification, the equations are:

$C=1$
$B+2C=0$
$A=B+C=0$

Solving simultaneously, the values are

$A=1$ and $B=-2$ and $C=1$

so that

$x^2/(x+1)^3=1/(x+1)^3-2/(x+1)^2+1/(x+1)$

have a nice day ! from the Philippines..

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