How are the remainder and factor theorems useful?

1 Answer
Amory W.
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(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.

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