17. Let $f(x)$ be a polynomial in $x$ of degree 5. When $f(x)$ is divided by $x-1, x-2, x-3$, $x-4$ and $x^{2}-x-1, \mathrm{f}(x)$ leaves a remainder of $3,1,7,36$ and $x-1$ respectively. Find the square of the remainder when $f(x)$ is divided by $x+1$.
17. Answer: 841
We have $\mathrm{f}(1)=3, \mathrm{f}(2)=1, \mathrm{f}(3)=7, \mathrm{f}(4)=36$ and
$$
f(x)=\left(x^{2}-x-1\right) g(x)+(x-1),
$$
where $\mathrm{g}(x)$ is a polynomial in $x$ of degree 3 . Hence $g(1)=-3, g(2)=0, g(3)=1$, and $g(4)=3$. Thus
$$
\begin{aligned}
f(x)= & \left(x^{2}-x-1\right)\left[(-3) \cdot \frac{(x-2)(x-3)(x-4)}{(-1)(-2)(-3)}+(1) \cdot \frac{(x-1)(x-2)(x-4)}{(2)(1)(-1)}\right. \\
& \left.+(3) \cdot \frac{(x-1)(x-2)(x-3)}{(3)(2)(1)}\right]+(x-1) .
\end{aligned}
$$

Thus, $f(-1)=-29$, so that its square is 841 .