Hi,

If you divide $f(x)$ by $x-2$ then the result is a polynomial I am going to call $p(x)$ and a remainder $R.$ The the degree of the remainder $R$ is less then the degree of the divisor $x-2$ so the remainder must be a constant. Hence we know that

\[f(x) = 4x^4-15x^2-4 = p(x) \times (x-2) + R.\]

Now substitute $x = 2$ into this equation to get

\[f(2) = p(2) \times (2 - 2) + R\]

and hence $f(2) = R.$

I hope this helps,
Harley