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.
