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 $A=-\dfrac{8}{25}$.