In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if f(x) is a polynomial, then x - a is a factor of f(x) if and only if f (a) = 0 (that is, a is a root of the polynomial). The theorem is a special case of the polynomial remainder theorem. The theorem results from basic properties of addition and multiplication. It follows that the theorem holds also when the coefficients and the element a belong to any commutative ring, and not just a field. In particular, since multivariate polynomials can be viewed as univariate in one of their variables, the following generalization holds : If and are multivariate polynomials and g is independent of X_1, then is a factor of if and only if is the zero polynomial.

Factorization of polynomials

Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent. The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows: Continuing the process until the polynomial f is factored completely, which all its factors is irreducible on or.

Example

Find the factors of Solution: Let p(x) be the above polynomial All possible factors of 2 are \pm 1 and \pm 2. Substituting x=-1, we get: So, (x-(-1)), i.e, (x+1) is a factor of p(x). On dividing p(x) by (x+1), we get Hence, Out of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic Thus the three irreducible factors of the original polynomial are x+1, and

Proofs

Several proofs of the theorem are presented here. If x-a is a factor of f(x), it is immediate that f(a)=0. So, only the converse will be proved in the following.

Proof 1

This proof begins by verifying the statement for a = 0. That is, it will show that for any polynomial f(x) for which f(0) = 0, there exists a polynomial g(x) such that. To that end, write f(x) explicitly as. Now observe that, so c_0 = 0. Thus,. This case is now proven. What remains is to prove the theorem for general a by reducing to the a = 0 case. To that end, observe that f(x + a) is a polynomial with a root at x = 0. By what has been shown above, it follows that for some polynomial g(x). Finally,.

Proof 2

First, observe that whenever x and y belong to any commutative ring (the same one) then the identity is true. This is shown by multiplying out the brackets. Let where R is any commutative ring. Write for a sequence of coefficients (c_i)_i. Assume f(a) = 0 for some a \in R. Observe then that. Observe that each summand has X - a as a factor by the factorisation of expressions of the form x^n - y^n that was discussed above. Thus, conclude that X - a is a factor of f(X).

Proof 3

The theorem may be proved using Euclidean division of polynomials: Perform a Euclidean division of f(x) by (x-a) to obtain where. Since, it follows that R is constant. Finally, observe that. So. The Euclidean division above is possible in every commutative ring since (x - a) is a monic polynomial, and, therefore, the polynomial long division algorithm does not involve any division of coefficients.

Corollary of other theorems

It is also a corollary of the polynomial remainder theorem, but conversely can be used to show it. When the polynomials are multivariate but the coefficients form an algebraically closed field, the Nullstellensatz is a significant and deep generalisation.