Example 3 Find the remainder when $f(x)=x^{1001}-1$ is divided by $p(x)=x^{4}+x^{3}+2 x^{2}+x+1$.
Analysis Let the remainder be $r(x)=a x^{3}+b x^{2}+c x+d, a, b, c, d \in \mathbf{R}$, then $f(x) \leftrightharpoons p(x) \cdot q(x)+$ $r(x)$, by considering the roots of $p(x)$, we can find $r(x)$.

Solution Let the remainder be $r(x)=a x^{3}+b x^{2}+c x+d, a, b, c, d \in \mathbf{R}$, factorize $p(x)$, we have $p(x)$ $=\left(x^{2}+1\right)\left(x^{2}+x+1\right)$, so $f(x)=\left(x^{2}+1\right)\left(x^{2}+x+1\right) q(x)+r(x)$
In (1), let $x=\mathrm{i}$ and $w\left(w=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$, since $p(\mathrm{i})=p(w)=0$, we have
$$
f(\mathrm{i})=r(\mathrm{i}), f(w)=r(w)
$$

And $f(\mathrm{i})=\mathrm{i}^{1001}-1=\mathrm{i}-1, r(\mathrm{i})=-a \mathrm{i}-b+c \mathrm{i}+d$,
$$
\begin{array}{l}
f(w)=w^{1001}-1=w^{2}-1=-\frac{3}{2}-\frac{\sqrt{3}}{2} \mathrm{i} \\
r(w)=a+b w^{2}+c w+d=a+d-\frac{b}{2}-\frac{c}{2}+\frac{\sqrt{3}}{2}(c-b) \mathrm{i} .
\end{array}
$$

Substitute them into (2), and by the necessary and sufficient condition for the equality of complex numbers, we get $\left\{\begin{array}{l}d-b=-1 \\ c-a=1 \\ a+b-\frac{b}{2}-\frac{c}{2}=-\frac{3}{2} \\ \frac{\sqrt{3}}{2}(c-b)=-\frac{\sqrt{3}}{2}\end{array}\right.$.
Solving these, we get $a=-1, b=1, c=d=0$. Therefore, the required remainder is $-x^{3}+x^{2}$.