Remainder Theorem: For a polynomial , the remainder of when dividing by is .
Proof: Dividing by , we obtain
where is the remainder. Since has degree 1, it follows that the remainder has degree 0, thus is a constant. Let . Substituting , we obtain . Thus, as claimed.
Factor Theorem: Let be a polynomial such that for some constant . Then is a factor of . Conversely, if is a factor of , then .
(Forward direction) Since , by the Remainder Theorem, we have . Hence, is a factor of .
(Backward direction) If is a factor of , then (by definition) the remainder when is divided by would be 0. By the Remainder Theorem, this is equal to . Hence, .
*Note: *There are no restrictions on the constant . It could be a real number, a complex number, or even a matrix!
The Remainder-Factor theorem is often used to help factorize polynomials without the use of long division. When combined with the Rational Roots theorem, this gives us a powerful factorization tool.
Completely factorize .
Solution: From the rational root theorem, we try numbers of the form , where divides 8 and divides 6. By the Remainder-Factor Theorem, we just need to calculate the values for which . Through trial and error, we obtain
This shows and are factors of , implying
Since the left hand side has degree 3, it follows that has degree 0, so is a constant we denote by . Substituting , we obtain . Therefore,
Application and Extensions
is a polynomial that leaves a remainder of 1 when divided by and leaves a remainder of 4 when divided by . What is the remainder when is divided by ?
We have . Since has degree 2, the remainder has degree at most 1, hence for some constants and . By the remainder factor theorem, we have and . Substituting and , we obtain
This implies , which has solution . Thus, the remainder when is divided by is .
In an attempt to discover a formula for the Fibonacci numbers, Alex finds a cubic polynomial such that , , and . What is the value of ?
Solution: Consider the cubic polynomial . Then and By the Remainder-Factor Theorem, we have , where is a polynomial. Since is a cubic, it follows that has degree 0, hence is a constant which we denote by . Substituting , we obtain
Note: The closed form of the Fibonacci sequence is an exponential function. This cannot be approximated using a polynomial function for large values of .
Suppose is a polynomial with integer coefficients and is an integer root of . Show that is also a polynomial with integer coefficients.
At first glance, it seems impossible to work directly from the Factor Theorem's result that .
Solution: Let the degree of be , and let
where are integers. By the remainder-factor theorem,
Using the fact that , we obtain
Since is integer combinations of polynomials with integer coefficients, it follows that is also a polynomial with integer coefficients.