partial fraction question 
$ \frac{125x^{2}+x+3}{x^{2}(x-5)} =
> \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-5)} | * x^{2}(x-5)$
$125x^{2}+x+3 = Ax(x-5) + B(x-5) + C (x^{2})$
$125x^{2}+x+3 = A x^{2} - 5Ax + Bx -5B +Cx^{2}$
$125x^{2}+x+3 = x^{2}(A+C) -x(A+B)-5B$
$3 = -5B \Rightarrow B = \frac{-3}{5}$
$-1 = A+B \Rightarrow A = -1 - B \Rightarrow A = \frac{-5}{5} - \frac{-3}{5} \Rightarrow A=\frac{-8}{5}$
$125 = A+C$

Where I did wrong in calculating of variable $A$, because correct answer is $A = \frac{-8}{25}$, but I get $A = \frac{-8}{5}$.
 A: HINT $\ $ It's simpler to use the Heaviside cover-up method. First, evaluating your $\rm\:E_2 = 2$nd equation at $\rm\:x = 0\:$ yields $\rm\:3 = -5\:b\:.\:$ Next, differentiating $\rm\:E_2\:$ and evaluating at $\rm\:x = 0\:$ yields $\rm\: 1 = b - 5\:a\:.$ Solve those for $\rm\:a,b\:$. Finally evaluating $\rm\:E_2\:$ at $\rm\:x=5\:$ yields $\rm\: 3133 = 25\:c\:.$ 
A: From
$$\frac{125x^{2}+x+3}{x^{2}(x-5)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-5}=
\frac{x(x-5)A+(x-5)B+x^{2}C}{x^{2}(x-5)}$$
it should be
$$125x^{2}+x+3=\left( A+C\right) x^{2}+\left( B-5A\right) x-5B$$
instead of
$$125x^{2}+x+3=\left( A+C\right) x^{2}+\left( A+B\right) x-5B.$$
Hence
$$\left\{ 
\begin{array}{c}
3=-5B \\ 
1=B-5A \\ 
125=A+C
\end{array}
\Leftrightarrow \right. \left\{ 
\begin{array}{c}
B=-\frac{3}{5} \\ 
1=-\frac{3}{5}-5A \\ 
125=A+C
\end{array}
\Leftrightarrow \right. \left\{ 
\begin{array}{c}
B=-\frac{3}{5} \\ 
A=-\frac{8}{25} \\ 
C=\frac{3133}{25}
\end{array}
\right. $$
and the expansion into partial fractions is $$\frac{125x^{2}+x+3}{x^{2}(x-5)}=-\frac{8}{25x}-\frac{3}{5x^{2}}+\frac{3133}{25\left( x-5\right) }.$$

Second method. Multiply 
$$\frac{125x^{2}+x+3}{x^{2}(x-5)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-5}
\qquad (\ast )$$
by $x-5$
$$\frac{125x^{2}+x+3}{x^{2}}=\frac{A(x-5)}{x}+\frac{B(x-5)}{x^{2}}+C$$
and let $x\rightarrow 5$
$$\lim_{x\rightarrow 5}\frac{125x^{2}+5+3}{x^{2}}=\frac{3133}{25}=C.$$
Multiply $(\ast )$ by $x^{2}$
$$\frac{125x^{2}+x+3}{x-5}=Ax+B+\frac{x^{2}}{x-5}C$$
and let $x\rightarrow 0$
$$\lim_{x\rightarrow 0}\frac{125x^{2}+x+3}{x-5}=-\frac{3}{5}=B.$$
Substitute $C$ and $B$ in $(\ast )$
$$\frac{125x^{2}+x+3}{x^{2}(x-5)}=\frac{A}{x}-\frac{3}{5}\frac{1}{x^{2}}+
\frac{3133}{25}\frac{1}{x-5}$$
and set, say, $x=1$
$$-\frac{125+1+3}{4}=A-\frac{3}{5}-\frac{3133}{25}\frac{1}{4},$$ 
to find $A=-\dfrac{8}{25}$.
A: The expression is of the form
$$\frac{Ax + b}{ax^2 + bx + c} + \frac{C}{x + d}$$ i.e Product of quadratic and Linear Factors.
Keeping the above form in mind,this is what i did.
$$\frac{125x^2 + x + 3}{x^2(x-5)} = \frac{Ax + B}{x^2} + \frac{C}{x-5}$$
Look for a common denominator in RHS
$$\frac{(Ax + B)(x - 5)}{x^2(x-5)} + \frac{c(x^2)}{(x-5)(x^2)}$$
Adding the above fraction
$$\frac{(Ax + B)(x - 5)+c(x^2)}{x^2(x - 5)}$$
Now equating the numerator from LHS with the numerter obtained above {RHS} as the denominator of both LHS and RHS contain the same expression,you obtain,
$$125x^2 + x + 3 = Ax^2 - 5Ax + Bx - 5B + cx^2$$
Let A = 5,find out the value of C,I'am pretty sure even B would get lost.Obtain the value of B and C by choosing a suitable value for x.
I'am having difficuilty in determining the value of B and C,Let me know if you could.
