How do you express $(x² + 2x-7 )/( (x+3)(x-1)²)$ in partial fractions?

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Answer 1 · 2016-11-29T11:43:30

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

Let's do the decomposition in partial fractions

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

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

Therefore,

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

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

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

Coefficients of $x^2$ , $=>$ , $1=A+B$

So, $B=1+1/4=5/4$

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

How do you express (x² + 2x-7 )/( (x+3)(x-1)²)(x² + 2x-7 )/( (x+3)(x-1)²) in partial fractions?