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
$$\frac{2x^{100}-3x+4}{x-1}$$

 A: If $r$ is the remainder of dividing $2x^{100}-3x+4$ by $x-1$, then $r$ is a constant and there exists a polynomial $q$ such that
$$p=2x^{100}-3x+4=q(x-1)+r,$$
which shows that $p(1)=r(1)=r$, so $r=3$.
A: It's a theorem that the remainder  on dividing a polynomial $p(x)$ by $x-\alpha$ is simply $p(\alpha)$.
A: The short, easy answer is that the remainder when you divide a polynomial $p(x)$ by $x-a$ is $p(a)$. This is a very useful fact, and not hard to prove.
However, even if you’ve forgotten this, brute force isn’t as slow as you suggest, provided that you keep your eyes open and spot the pattern. The first stage of an ordinary long division gives you a partial quotient of $2x^{99}$ and a remainder of $2x^{99}-3x+4$. The next stage extends the partial quotient to $2x^{99}+2x^{98}$ and leaves you a remainder of $2x^{98}-3x+4$. Clearly this is going to continue until at some point you have a partial quotient of
$$2x^{99}+2x^{98}+\ldots+2x^2$$
and a remainder of $2x^2-3x+4=(x-1)(2x-1)+3$.
A: 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:
$$=2\left(\sum_{n=1}^{100}{100\choose n}(x-1)^{n-1}\right)-3+\frac{3}{x-1}$$
$$=2\left(\sum_{n=1}^{100}{100\choose n}\sum_{i=0}^{n-1}(-1)^ix^{i}\right)-3+\frac{3}{x-1}$$
$$=2\sum_{n=0}^{99}x^n-3+\frac{3}{x-1}$$
$$=2\sum_{n=1}^{99}x^n-1+\frac{3}{x-1}$$
