Factor theorem

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:^[3]

1. Deduce the candidate of zero ${\displaystyle a}$ of the polynomial ${\displaystyle f}$ from its leading coefficient ${\displaystyle a_{n}}$ and constant term ${\displaystyle a_{0}}$. (See Rational root theorem.)
2. Use the factor theorem to conclude that ${\displaystyle (x-a)}$ is a factor of ${\displaystyle f(x)}$.
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 ${\displaystyle x\neq a}$ of ${\displaystyle f(x)=0}$ is a root of ${\displaystyle g(x)=0}$. Since the polynomial degree of ${\displaystyle g}$ is one less than that of ${\displaystyle f}$, it is "simpler" to find the remaining zeros by studying ${\displaystyle g}$.

Continuing the process until the polynomial ${\displaystyle f}$ is factored completely, which all its factors is irreducible on ${\displaystyle \mathbb {R} [x]}$ or ${\displaystyle \mathbb {C} [x]}$.

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

Solution: Let ${\displaystyle p(x)}$ be the above polynomial

Constant term = 2

Coefficient of ${\displaystyle x^{3}=1}$

All possible factors of 2 are ${\displaystyle \pm 1}$ and ${\displaystyle \pm 2}$. Substituting ${\displaystyle x=-1}$, we get:

${\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2=0}$

So, ${\displaystyle (x-(-1))}$, i.e, ${\displaystyle (x+1)}$ is a factor of ${\displaystyle p(x)}$. On dividing ${\displaystyle p(x)}$ by ${\displaystyle (x+1)}$, we get

Quotient = ${\displaystyle x^{2}+6x+2}$

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

Out of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic ${\displaystyle -3\pm {\sqrt {7}}.}$ Thus the three irreducible factors of the original polynomial are ${\displaystyle x+1,}$ ${\displaystyle x-(-3+{\sqrt {7}}),}$ and ${\displaystyle x-(-3-{\sqrt {7}}).}$

Several proofs of the theorem are presented here.

If ${\displaystyle x-a}$ is a factor of ${\displaystyle f(x),}$ it is immediate that ${\displaystyle f(a)=0.}$ So, only the converse will be proved in the following.

This proof begins by verifying the statement for ${\displaystyle a=0}$. That is, it will show that for any polynomial ${\displaystyle f(x)}$ for which ${\displaystyle f(0)=0}$, there exists a polynomial ${\displaystyle g(x)}$ such that ${\displaystyle f(x)=x\cdot g(x)}$. To that end, write ${\displaystyle f(x)}$ explicitly as ${\displaystyle c_{0}+c_{1}x^{1}+\dotsc +c_{n}x^{n}}$. Now observe that ${\displaystyle 0=f(0)=c_{0}}$, so ${\displaystyle c_{0}=0}$. Thus, ${\displaystyle f(x)=x(c_{1}+c_{2}x^{1}+\dotsc +c_{n}x^{n-1})=x\cdot g(x)}$. This case is now proven.

What remains is to prove the theorem for general ${\displaystyle a}$ by reducing to the ${\displaystyle a=0}$ case. To that end, observe that ${\displaystyle f(x+a)}$ is a polynomial with a root at ${\displaystyle x=0}$. By what has been shown above, it follows that ${\displaystyle f(x+a)=x\cdot g(x)}$ for some polynomial ${\displaystyle g(x)}$. Finally, ${\displaystyle f(x)=f((x-a)+a)=(x-a)\cdot g(x-a)}$.

First, observe that whenever ${\displaystyle x}$ and ${\displaystyle y}$ belong to any commutative ring (the same one) then the identity ${\displaystyle x^{n}-y^{n}=(x-y)(y^{n-1}+x^{1}y^{n-2}+\dotsc +x^{n-2}y^{1}+x^{n-1})}$ is true. This is shown by multiplying out the brackets.

Let ${\displaystyle f(X)\in R\left[X\right]}$ where ${\displaystyle R}$ is any commutative ring. Write ${\displaystyle f(X)=\sum _{i}c_{i}X^{i}}$ for a sequence of coefficients ${\displaystyle (c_{i})_{i}}$. Assume ${\displaystyle f(a)=0}$ for some ${\displaystyle a\in R}$. Observe then that ${\displaystyle f(X)=f(X)-f(a)=\sum _{i}c_{i}(X^{i}-a^{i})}$. Observe that each summand has ${\displaystyle X-a}$ as a factor by the factorisation of expressions of the form ${\displaystyle x^{n}-y^{n}}$ that was discussed above. Thus, conclude that ${\displaystyle X-a}$ is a factor of ${\displaystyle f(X)}$.

The theorem may be proved using Euclidean division of polynomials: Perform a Euclidean division of ${\displaystyle f(x)}$ by ${\displaystyle (x-a)}$ to obtain ${\displaystyle f(x)=(x-a)Q(x)+R(x)}$ where ${\displaystyle \deg(R)<\deg(x-a)}$. Since ${\displaystyle \deg(R)<\deg(x-a)}$, it follows that ${\displaystyle R}$ is constant. Finally, observe that ${\displaystyle 0=f(a)=R}$. So ${\displaystyle f(x)=(x-a)Q(x)}$.

The Euclidean division above is possible in every commutative ring since ${\displaystyle (x-a)}$ is a monic polynomial, and, therefore, the polynomial long division algorithm does not involve any division of coefficients.

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.