How can I find the remainder of this polynomial division? This is for algebra 2 honors. Of course I can long divide or use synthetic, but that would take a while.

Determine the remainder of
$$\frac{2x^{100}-3x+4}{x-1}$$


The remainder theorem says the remainder of $f(x)$ when divided by $(x-a)$ is $f(a)$ so the answer would be: $2\times1^{100}-3x+4=3$.
If instead you wanted to work out the quotient that is a bit harder. If you let $t=x-1$ then you can rewrite it as:
$$\frac{2(t+1)^{100}-3(t+1)+4}{t}=\frac{2\left(\sum_{n=0}^{100}{100\choose n}t^n\right)-3t+1}{t}$$
$$=2\left(\sum_{n=1}^{100}{100\choose n}t^{n-1}\right)-3+\frac{3}{t}$$
Substituting back for $x$ will then give:
$$=2\left(\sum_{n=1}^{100}{100\choose n}(x-1)^{n-1}\right)-3+\frac{3}{x-1}$$
$$=2\left(\sum_{n=1}^{100}{100\choose n}\sum_{i=0}^{n-1}(-1)^ix^{i}\right)-3+\frac{3}{x-1}$$
$$=2\sum_{n=0}^{99}x^n-3+\frac{3}{x-1}$$
$$=2\sum_{n=1}^{99}x^n-1+\frac{3}{x-1}$$
