Factor theorem · Lablypedia

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 $f (X_{1}, \dots, X_{n})$ and $g (X_{2}, \dots, X_{n})$ are multivariate polynomials and $g$ is independent of $X_{1}$ , then $X_{1} - g (X_{2}, \dots, X_{n})$ is a factor of $f (X_{1}, \dots, X_{n})$ if and only if $f (g (X_{2}, \dots, X_{n}), X_{2}, \dots, X_{n})$ 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:

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.)
Use the factor theorem to conclude that $(x - a)$ is a factor of $f (x)$ .
Compute the polynomial $g (x) = \frac{f (x)}{(x - a)}$ , for example using polynomial long division or synthetic division.
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 $R [x]$ or $C [x]$ .

Example

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

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} + 6 x + 2

Hence, $p (x) = (x^{2} + 6 x + 2) (x + 1)$

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

One more chapter is free with a Lably account.

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 $c_{0} + c_{1} x^{1} + \dots + c_{n} x^{n}$ . Now observe that $0 = f (0) = c_{0}$ , so $c_{0} = 0$ . Thus, $f (x) = x (c_{1} + c_{2} x^{1} + \dots + c_{n} x^{n - 1}) = x \cdot g (x)$ . 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 $x^{n} - y^{n} = (x - y) (y^{n - 1} + x^{1} y^{n - 2} + \dots + x^{n - 2} y^{1} + x^{n - 1})$ is true. This is shown by multiplying out the brackets.

Let $f (X) \in R [X]$ 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 $\deg (R) < \deg (x - a)$ . 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.