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

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Answer 1 · 2018-08-07T07:30:03

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

Perform the decomposition into partial fractions as follows

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

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

The denominators are the same, compare the numerators

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

Let $x=2$ , $=>$ , $4-4+1=A$ , $=>$ , $A=1$

Coefficients of $x^2$

$1=C$

Coefficients of $x$

$-2=B-4C$

$B=4C-2=2$

Finally,

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

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