Remainder Theorem vs. Factor Theorem Is it right to say that the remainder theorem and the factor theorem are similar, but not the same processes? For context, this was a question on a homework sheet provided to me by my advanced functions teacher. The question asked for a way other than remainder theorem to determine if a binomial was a factor of a polynomial. I argued that factor theorem was a way, but she said that the two were basically the same and that synthetic division would be the correct answer. My train of thought was:
Remainder Theorem - for any polynomial f(x), if you divide it by the binomial x−a, the remainder is equal to the value of f(a).
Factor Theorem - if a is a zero of a polynomial f(x), then (x−a) is a factor of f(x)
I concluded, with the help of an online resource, that the remainder theorem links the remainder of division by a binomial with the value of a function at a point, while the factor theorem links the factors of a polynomial to its zeros. While they may seem the same, mathematically, they are different, right? That's why they are two distinct theorems and not one. I also thought that if the "remainder theorem and factor theorem are basically the same" then the remainder theorem and synthetic division are also the same because you are using the rational zeroes theorem in both cases.
Any mathematical or logical help to prove the difference between the remainder theorem and the factor theorem would be greatly appreciated. Thanks
 A: The factor theorem is just a special case of the remainder theorem. They are distinct only because the remainder theorem is more general. But using the factor theorem does indeed count as using the remainder theorem.
if you want to show that $(x-a)$ is a factor of $f(x)$ without using the remainder theorem, the way to do it would be to first divide $f(x)$ by $(x-a)$ and then rather than using the fact that the remainder is 0, you just use the quotient polynomial that you got, and show that $f(x) = (x-a) \times \frac{f(x)}{(x-a)} $ (where rather than writing $\frac{f(x)}{(x-a)}$  you would write what ever polynomial you got when doing the division.
To be clear about your question though, from a purely logical point of view, lets look at what the statements are. The remainder theorem says that the remainder of a function $f(x)$ when divided by $(x-a)$ is equal to $f(a)$.
To for $(x-a)$ to be a factor of $f(x)$ by definition means that $(x-a)$ divides $f(x)$ which means that the remainder is 0. So in particular by the factor theorem we can conclude that $(x-a)$ is a factor of $f(x)$ whenever $f(a)=0$. We call this result the factor theorem.
Notice, that in the proof of the factor theorem, we used the remainder theorem. If we did not already know the remainder theorem, we would not know the factor theorem either, since its proof relies on the remainder theorem. (It is possible to prove the factor theorem independently of the remainder theorem, but in your class you have not done that, so the point sticks). Now anytime you use the factor theorem, your result only holds because of the proof that you have come up with. But that proof uses the remainder theorem. So using the factor theorem implies using the remainder theorem.
You are right that they are not equivalent. Because using the remainder theorem does not imply using the factor theorem. However them being equivalent is not necessary for the factor theorem to be disallowed in this question. The fact that the factor theorem implies using the remainder theorem is clearly already enough to cause issue.
