# How are the remainder and factor theorems useful?

Aug 14, 2014

You have only learned tools to factor quadratics; for higher degree polynomials, you can use long or synthetic division.

However, long/synthetic division is slow. If you guess the wrong divisor/factor, you will be wasting even more time.

For example: P(x)=2x^2−x−1. (I am using a quadratic for the sake of brevity).

Using the remainder theorem, we can quickly get the remainder with P(2)=2⋅2^2−2−1=8−2−1=5 and P(0)=2⋅0^2−0−1=−1. So, $x - 2$ and $x - 0$ are not factors of $P \left(x\right)$. 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. P(1)=2⋅1^2−1−1=0, so x−1 is a factor of $P \left(x\right)$ by the factor theorem.

Once you have a factor, you can proceed with long/synthetic division to get your quotient. Then you will need to repeat and factor the quotient. Again, if it is a quadratic, switch to your quadratic factoring methods.