How do you express $(5x^2+3x-2)/(x^3+2x^2)$ in partial fractions?

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Answer 1 · 2016-10-27T08:54:51

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

We need to simplify the denominator
$x^3+2x^2=x^2(x+2)$

Let $(5x^2+3x-2)/(x^3+2x^2)=(5x^2+3x-2)/(x^2(x+2))=A/x^2+B/x+C/(x+2)$
$=(A(x+2)+Bx+Cx^2)/(x^2(x+2))$
So $5x^2+3x-2=A(x+2)+Bx(x+2)+Cx^2$

Let $x=0$ then $-2=2A$ $=>$ $A=-1$

Coefficients of $x^2$
$5=B+C$

Coefficients of $x$
$3=A+2B$ $=>$ $3=-1+2B$ $=>$ $B=2$

and $5=2+C$ $=>$ $C=3$

so the result is
$(5x^2+3x-2)/(x^3+2x^2)=-1/x^2+2/x+3/(x+2)$

How do you express (5x^2+3x-2)/(x^3+2x^2) (5x^2+3x-2)/(x^3+2x^2) in partial fractions?