5. [20] The curves $y=x^{2}(x-3)^{2}$ and $y=\left(x^{2}-1\right)(x-2)$ intersect at a number of points in the real plane. Determine the sum of the $x$-coordinates of these points of intersection.
Answer: 7. Because the first curve touches the $x$-axis at $x=0$ and $x=3$ while the second curve crosses the $x$-axis at $x= \pm 1$ and $x=2$, there are four points of intersection. In particular, the points of intersection have $x$-coordinates determined by the difference of the two curves:
$$
0=x^{2}(x-3)^{2}-\left(x^{2}-1\right)(x-2)=\left(x^{4}-6 x^{3}+\cdots\right)-\left(x^{3}+\cdots\right)=x^{4}-7 x^{3}+\cdots .
$$

We need only the first two coefficients to determine $x_{1}+x_{2}+x_{3}+x_{4}=-\left(\frac{-7}{1}\right)=7$.