How do you write the partial fraction decomposition of the rational expression $(x² + 2x-7) /( (x+3)(x-1)²)$ ?

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Answer 1 · 2017-02-28T12:05:04

The answer is $=(-1/4)/(x+3)-1/(x-1)^2+(5/4)/(x-1)$

Let 's perform the decomposition into partial fractions

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

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

As the denominators are the same, we compare the numerators

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

Let $x=-3$ , $=>$ , $-4=16A$ , $=>$ , $A=-1/4$

Let $x=1$ , $=>$ , $-4=4B$ , $=>$ , $B=-1$

Coefficients of $x^2$

$1=A+C$ , $=>$ , $C=1-A=1+1/4=5/4$

Therefore,

$(x^2+2x-7)/((x+3)(x-1)^2)=(-1/4)/(x+3)-1/(x-1)^2+(5/4)/(x-1)$

How do you write the partial fraction decomposition of the rational expression (x² + 2x-7) /( (x+3)(x-1)²)(x² + 2x-7) /( (x+3)(x-1)²)?