因式定理

该定理指出，一个多项式 f(x) 有一个因式 (ax-b) 当且仅当 f(b/a)=0。

多项式的因式分解

因式定理普遍应用于找到一个多项式的因式或多项式方程的根的两类问题。从定理的推论结果，这些问题基本上是等价的。

若多项式已知一个或数个零点，因式定理也可以移除多项式中已知零点的部分，变成一个阶数较低的多项式，其零点即为原多项式中剩下的零点，以简化多项式求根的过程。方法如下：

先设法找出多项式 f 的一个零点 a。

利用因式定理确认 (x-a) 是多项式 f(x) 的因式。

利用长除法计算多项式 g(x)=(f(x))/(x-a)。 f(x)=0 中，所有满足 x≠ a 条件的根 x 都是方程式 g(x)=0 的根。因为 g(x) 的多项式阶数较 f(x) 要小。因此要找出多项式 g 的零点可能会比较简单。

欲使A=BQ+R成立,就令除式BQ=0,则被除式A=R能使此方程式成立,则被除式=(商式)(除式)+余式或被除式/除式=商式+余式/除式。

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(X₁,…,Xₙ) and g(X₂,…,Xₙ) are multivariate polynomials and g is independent of X₁, then X₁-g(X₂,…,Xₙ) is a factor of f(X₁,…,Xₙ) if and only if f(g(X₂,…,Xₙ),X₂,…,Xₙ) 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:

Deduce the candidate of zero a of the polynomial f from its leading coefficient aₙ and constant term a₀. (See Rational root theorem.)

Use the factor theorem to conclude that (x-a) is a factor of f(x).

Compute the polynomial g(x)=(f(x))/((x-a)), for example using polynomial long division or synthetic division.

Conclude that any root x≠ a 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 R [x] C [x].

Example

Find the factors of x³+7x²+8x+2.

• Solution: Let p(x) be the above polynomial
• Constant term = 2
• Coefficient of x³=1

All possible factors of 2 are ± 1 and ± 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) (x+1), we get
• Quotient = x²+6x+2
• Hence, p(x)=(x²+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± √7. Thus the three irreducible factors of the original polynomial are x+1,

x-(-3+√7), and x-(-3-√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· g(x). To that end, write f(x) explicitly as （公式略）. Now observe that 0=f(0)=c₀, so c₀=0. Thus, （公式略）. 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· g(x) for some polynomial g(x). Finally, f(x)=f((x-a)+a)=(x-a)· g(x-a).

Proof 2

First, observe that whenever x and y belong to any commutative ring (the same one) then the identity （公式略） is true. This is shown by multiplying out the brackets.

Let f(X)∈ R[X] where R is any commutative ring. Write f(X)=∑ ᵢcᵢXⁱ for a sequence of coefficients (cᵢ)ᵢ. Assume f(a)=0 for some a∈ R. Observe then that f(X)=f(X)-f(a)=∑ ᵢcᵢ(Xⁱ-aⁱ). Observe that each summand has X-a as a factor by the factorisation of expressions of the form xⁿ-yⁿ 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) (x-a) to obtain f(x)=(x-a)Q(x)+R(x) where deg(R)<deg(x-a). Since deg(R)<deg(x-a), 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.