Factor theorem

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 $$f(X_1,\ldots,X_n)$$ 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
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 $$\mathbb{R}[x]$$ or $$\mathbb{C}[x]$$.
 * 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 $$g$$.

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 $$x-(-3-\sqrt{7}).$$

Proof
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 argument begins by verifying the theorem for $$a = 0$$. That is, it aims to show that for any polynomial $$f(x)$$ for which $$f(0) = 0$$ it is true that $$f(x) =x\cdot g(x)$$ for some polynomial $$g(x)$$. 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 $$ f(x) = (x - a) Q + R$$ 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 $$f(x) = (x - a)Q$$.

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 involves 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.