Polynomial Remainder Theorem
We use Euclidean polynomial division with dividend and divisor . The result states that there exists a quotient and remainder such that with . We wish to show that is equal to the constant . Because , . Hence, is a constant, . Plugging this into our original equation and rearranging a bit yields After substituting into this equation, we deduce that ; thus, the remainder upon diving by is equal to , as desired.
The strategy used in the above proof can be generalized to divisors with degree greater than . A more general method, with any dividend and divisor , is to write , and then substitute the zeroes of to eliminate and find values of . Example 2 showcases this strategy.
Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.
What is the remainder when is divided by ?
Solution: Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that , and . A common mistake is to forget to flip the negative sign and assume , but simplifying the linear equation yields . Thus, the answer is , or , which is equal to . .