# How do I use the remainder theorem to evaluate polynomials?

The remainder theorem only applies if your divisor is a monic linear binomial, that is, $x - a$. If you have a polynomial $P \left(x\right)$ and divide it by $x - a$, then the remainder is $P \left(a\right)$. 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: $P \left(x\right) = 2 {x}^{2} - x - 1$ divided by $x - 2$. If we do long or synthetic division, we get $Q \left(x\right) = 2 x + 3$ and $R \left(x\right) = 5$. But using the remainder theorem, we can quickly get the remainder with $P \left(2\right) = 2 \cdot {2}^{2} - 2 - 1 = 8 - 2 - 1 = 5$.
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 $P \left(x\right)$. But $P \left(1\right) = 2 \cdot {1}^{2} - 1 - 1 = 0$, so $x - 1$ is a factor of $P \left(x\right)$.
If instead, we tried $P \left(0\right) = 2 \cdot {0}^{2} - 0 - 1 = - 1$, so $x - 0$ is not a factor. But consider that $P \left(0\right)$ is below the x-axis and $P \left(2\right)$ is above the x-axis; this means the $P \left(x\right)$ must cross the x-axis somewhere between 0 and 2. This should lead you to try $P \left(1\right)$ as a factor.