How do you write the partial fraction decomposition of the rational expression $(x^3-x^2+1) / (x^4-x^3)$ ?

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Answer 1 · 2015-12-23T23:11:54

$(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1$

$(x^3-x^2+1) / (x^4-x^3) = (x^3-x^2)/ (x^4-x^3) + 1/(x^4-x^3)$

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

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

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

now focus on $1/(x^3(x-1))$

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

Multiply both side by $x^3(x-1)$

$1 = A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^3$

$1=Ax-A+Bx^2-Bx+Cx^3-Cx^2+Dx^3$

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

$C+D = 0$
$B-C = 0$
$A-B = 0$
$A = -1$

Just by looking we have

$A = - 1$
$B = -1$
$C = -1$
$D = 1$

So

$(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1)$

How do you write the partial fraction decomposition of the rational expression (x^3-x^2+1) / (x^4-x^3) (x^3-x^2+1) / (x^4-x^3)?