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