$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5 \
& \lim _{x \rightarrow 0} \frac{\left.a x^4+b x^3+c x^2+d x+e\right)}{x^2}=5 \
& c=5 \text { and } d=e=0 \
& f(x)=a x^4+b x^3+5 x^2 \
& f^{\prime}(x)=4 a x^3+3 b x^2+10 x \
& =x\left(4 a x^2+3 b x+10\right)
\end{aligned}$
has extremes at 4 and so f′(4)=0&f′(5)=0
so a=81& b=2−3
so f(2)=81×24−23×23+5×22
=2−12+20=10