How do you express $(7x-2)/((x-3)^2(x+1))$ in partial fractions?

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Answer 1 · 2017-02-11T07:39:53

The answer is $=(19/4)/(x-3)^2+(9/16)/(x-3)+(-9/16)/(x+1)$

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let $x=3$ , $=>$ , $19=4A$ , $=>$ , $A=19/4$

Let $x=-1$ , $=>$ , $-9=16C$ , $=>$ , $C=-9/16$

Coefficients of $x^2$

$0=B+C$ , $=>$ , $B=-C=9/16$

Therefore,

$(7x-2)/((x-3)^2(x+1))=(19/4)/(x-3)^2+(9/16)/(x-3)+(-9/16)/(x+1)$

How do you express (7x-2)/((x-3)^2(x+1)) (7x-2)/((x-3)^2(x+1)) in partial fractions?