Your question isn't phrased quite correctly. The remainder theorem is a short cut to find the remainder of polynomial long division or synthetic division.
The remainder theorem only applies if your divisor is a monic linear binomial, that is, x-a. If you have a polynomial P(x) and divide it by x-a, then the remainder is P(a). Note that the remainder theorem doesn't give you the quotient, so you can't use it for questions that are looking for the quotient and remainder.
For example: P(x)=2x^2-x-1 divided by x-2. If we do long or synthetic division, we get Q(x)=2x+3 and R(x)=5. But using the remainder theorem, we can quickly get the remainder with P(2)=2*2^2-2-1=8-2-1=5.
When we combine the remainder theorem with the factor theorem, we can use it to find/verify the factors of the polynomial. So, x-2 is not a factor of P(x). But P(1)=2*1^2-1-1=0, so x-1 is a factor of P(x).
If instead, we tried P(0)=2*0^2-0-1=-1, so x-0 is not a factor. But consider that P(0) is below the x-axis and P(2) is above the x-axis; this means the P(x) must cross the x-axis somewhere between 0 and 2. This should lead you to try P(1) as a factor.
