Factor theorem

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title: "Factor theorem" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["theorems-about-polynomials"] description: "Polynomial zeros related to linear factors" topic_path: "general/theorems-about-polynomials" source: "https://en.wikipedia.org/wiki/Factor_theorem" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

::summary Polynomial zeros related to linear factors ::

::figure[src="https://upload.wikimedia.org/wikipedia/commons/f/fe/Polinomio_de_grado_4.png"] ::

In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if f(x) is a (univariate) 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 f(X_1,\ldots,X_n) and g(X_2, \ldots,X_n) are multivariate polynomials and g is independent of X_1, then X_1 - g(X_2, \ldots,X_n) is a factor of f(X_1,\ldots,X_n) if and only if f(g(X_2, \ldots,X_n),X_2, \ldots,X_n) is the zero polynomial.

Factorization of polynomials

Main article: 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:

1. Deduce the candidate of zero a of the polynomial f from its leading coefficient a_n and constant term a_0. (See Rational root theorem.)
2. Use the factor theorem to conclude that (x-a) is a factor of f(x).
3. Compute the polynomial g(x) = \dfrac{f(x)}{(x-a)} , for example using polynomial long division or synthetic division.
4. Conclude that any root x \neq a of f(x)=0 is a root of g(x)=0. Since the polynomial degree of g is one less than that of f, it is "simpler" to find the remaining zeros by studying g. Continuing the process until the polynomial f is factored completely, which all its factors is irreducible on \mathbb{R}[x] or \mathbb{C}[x].

Example

Find the factors of x^3 + 7x^2 + 8x + 2.

Solution: Let p(x) be the above polynomial :Constant term = 2 : Coefficient of x^3=1 All possible factors of 2 are \pm 1 and \pm 2 . Substituting x=-1, we get: :(-1)^3 + 7(-1)^2 + 8(-1) + 2 = 0 So, (x-(-1)), i.e, (x+1) is a factor of p(x). On dividing p(x) by (x+1), we get : Quotient = x^2 + 6x + 2 Hence, p(x)=(x^2 + 6x + 2)(x+1)

Out of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic -3\pm \sqrt{7}. Thus the three irreducible factors of the original polynomial are x+1, x-(-3+\sqrt{7}), and x-(-3-\sqrt{7}).

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 f(x) =x\cdot g(x). To that end, write f(x) explicitly as c_0 +c_1 x^1 + \dotsc + c_n x^n. Now observe that 0 = f(0) = c_0, so c_0 = 0. Thus, f(x) = x(c_1 + c_2 x^1 + \dotsc + c_{n} x^{n-1}) = x \cdot g(x). 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 f(x + a) = x \cdot g(x) for some polynomial g(x). Finally, f(x) = f((x - a) + a) = (x - a)\cdot g(x - a).

Proof 2

First, observe that whenever x and y belong to any commutative ring (the same one) then the identity x^n - y^n = (x - y)(y^{n-1} + x^1 y^{n-2} + \dotsc + x^{n-2}y^{1} + x^{n-1}) is true. This is shown by multiplying out the brackets.

Let f(X) \in R\left[ X \right] where R is any commutative ring. Write f(X) = \sum_i c_i X^i for a sequence of coefficients (c_i)i. Assume f(a) = 0 for some a \in R. Observe then that f(X) = f(X) - f(a) = \sum{i} c_i(X^i - a^i). 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 f(x) = (x - a) Q(x)+ R(x) where \deg(R) . Since \deg(R) , it follows that R is constant. Finally, observe that 0 = f(a) = R. So f(x) = (x - a)Q(x) .

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.

References

References

1. Sullivan, Michael. (1996). "Algebra and Trigonometry". Prentice Hall.
2. (September 2009). "Longman ICSE Mathematics Class 10". Dorling Kindersley (India).
3. Bansal, R. K.. "Comprehensive Mathematics IX". Laxmi Publications.

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