Factor theorem

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 $f(X_1,\dots ,X_n)$ 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

Template:Main 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 $g(x)=\frac{f(x)}{(x-a)}$, for example using polynomial long division or synthetic division.
4. Conclude that any root $x\ne a$ 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 $x^3+7x^2+8x+2.$

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 $c_0+c_1{x}^1+\dots +c_n{x}^n$. 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 $f(X)\in R\left[X\right]$ 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 $\mathrm{deg}(R)<\mathrm{deg}(x-a)$. Since $\mathrm{deg}(R)<\mathrm{deg}(x-a)$, 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.