The remainder factor theorem is actually two theorems that relate the roots of a polynomial with its linear factors. The theorem is often used to help factorize polynomials without the use of long division. Especially when combined with the rational root theorem, this gives us a powerful tool to factor polynomials.
For a polynomial , the remainder of upon division by is .
Let be a polynomial such that for some constant . Then is a factor of . Conversely, if is a factor of , then .
Completely factorize .
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 right hand side has degree 3, it follows that has degree 0, so it is a constant we denote by . Substituting , we obtain . Therefore,
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, and 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
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 and thus 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 .
Let the degree of be , and let
where are integers. Then 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.
Dividing by , we obtain
where is the remainder. Since has degree 1, it follows that the remainder has degree 0 and thus is a constant. Let . Then substituting , we obtain . Thus, the remainder is , as claimed.
Since , by the remainder theorem, we have . Hence, is a factor of .
If is a factor of , then (by definition) the remainder of upon division 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!