Question

How are the remainder and factor theorems useful?

Answer 1

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(x)`. 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. `P(1)=2⋅1^2−1−1=0`, so `x−1` is a factor of `P(x)` 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.