Factor theorem

In algebra, the factor theorem connects polynomial factors with their zeros. Specifically, if $f(x)$ is a polynomial, then $f(a)=0$ if and only if $x-a$ is a factor of $f(x)$. An $a$ for which $f(a)=0$ is a called a root. The theorem is a special case of the polynomial remainder theorem.

This theorem is commonly applied to univariate polynomials where both the coefficients and the root $a$ are real or complex numbers.

From an abstract algebra point of view, it's relevant that some proofs of the theorem depend only on basic addition and multiplication. From this, it follows that the theorem applies even when the coefficients and root $a$ are part of any commutative ring, and not just a field. In particular, this implies a generalisation of the theorem from univariate to multivariate polynomials (since multivariate polynomials can be thought of as univariate in one their variables):

 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 zero.

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 ${\textstyle g(x)={\dfrac {f(x)}{(x-a)}}}$, 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 $x-(-3-{\sqrt {7}}).$

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, $f(x)=f((x-a)+a)=(x-a)\cdot g(x-a)$.

Remark: The proof implicitly uses the associativity property of polynomial composition along with the fact that $x$ is the unit for polynomial composition. This is being made explicit because in abstract algebra polynomials are not defined as functions, and so these properties of polynomial composition are not immediate.

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 $f(X)$.

As a corollary of other theorems

The theorem may be proved using Euclidean division of polynomials. 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.