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 $A = \frac{-8}{5}$.

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.
