How do you write the partial fraction decomposition of the rational expression $(3x^2 -7x+1) / (x-1)^3$ ?

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Answer 1 · 2016-11-21T06:36:52

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

The decomposition in partial fractions is

$(3x^2-7x+1)/(x-1)^3=A/(x-1)^3+B/(x-1)^2+C/(x-1)$

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

Therefore,

$(3x^2-7x+1)=A+B(x-1)+C(x-1)^2$

Let $x=1$ , $=>$ , $-3=A$

$A+B+C=1$

Coefficients of $x^2$
$3=C$

Coefficients of $x$

$-7=-B-2C$ , $=>$ , $B=-1$

so, $(3x^2-7x+1)/(x-1)^3=(-3)/(x-1)^3-1/(x-1)^2+3/(x-1)$

How do you write the partial fraction decomposition of the rational expression (3x^2 -7x+1) / (x-1)^3(3x^2 -7x+1) / (x-1)^3?