5. Consider the two curves $y=2 x^{3}+6 x+1$ and $y=-\frac{3}{x^{2}}$ in the Cartesian plane. Find the number of distinct points at which these two curves intersect.
(A) 1
(B) 2
(C) 3
(D) 0
(E) 5
5. Ans: A
To find the intersection points of the two curves, we solve their equations simultaneously.
$$
\left\{\begin{array}{l}
y=2 x^{3}+6 x+1, \\
y=-\frac{3}{x^{2}} .
\end{array}\right.
$$

Thus, we have
$$
\begin{aligned}
2 x^{3}+6 x+1=-\frac{3}{x^{2}} & \Longrightarrow 2 x^{5}+6 x^{3}+x^{2}+3=0 \\
& \Longrightarrow\left(2 x^{3}+1\right)\left(x^{2}+3\right)=0 \\
& \Longrightarrow 2 x^{3}+1=0, \text { since } x^{2}+3>0 \\
& \Longrightarrow x^{3}=-\frac{1}{2} \\
& \Longrightarrow x=-\frac{1}{\sqrt[3]{2}} \\
& \Longrightarrow y=-\frac{3}{\sqrt[3]{4}} .
\end{aligned}
$$

Thus, the only point of intersection is $\left(-\frac{1}{\sqrt[3]{2}},-\frac{3}{\sqrt[3]{4}}\right)$.