When the polynomial $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$ is divided by each of $x-4, x-3, x+3$, and $x+4$, the remainder in each case is 102. Determine all values of $x$ for which $f(x)=246$.
When $f(x)$ is divided by $x-4$, the remainder is 102 .

This means that $f(x)-102$ has a factor of $x-4$.

Similarly, $f(x)-102$ is divisible by each of $x-3, x+3$ and $x+4$.

Since $f(x)$ is a quartic polynomial, then $f(x)-102$ is a quartic polynomial and so can have at most four distinct linear factors.

Since the leading coefficient of $f(x)$ is 1 , then the leading coefficient of $f(x)-102$ is also 1 .

This means that $f(x)-102=(x-4)(x-3)(x+3)(x+4)$.

Thus, $f(x)=(x-4)(x-3)(x+3)(x+4)+102$.

Therefore, the following equations are equivalent:

$$
\begin{aligned}
f(x) & =246 \\
(x-4)(x-3)(x+3)(x+4)+102 & =246 \\
(x-4)(x+4)(x-3)(x+3) & =144 \\
\left(x^{2}-16\right)\left(x^{2}-9\right) & =144 \\
x^{4}-25 x^{2}+144 & =144 \\
x^{4}-25 x^{2} & =0 \\
x^{2}\left(x^{2}-25\right) & =0 \\
x^{2}(x-5)(x+5) & =0
\end{aligned}
$$

Thus, the values of $x$ for which $f(x)=246$ are $x=0,5,-5$.