Example

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

Proof 1

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

Proof 2

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 ${\displaystyle f(X)}$.

Proof 3

The theorem may be proved using Euclidean division of polynomials: Perform a Euclidean division of ${\displaystyle f(x)}$ by ${\displaystyle x-a}$ to obtain ${\displaystyle f(x)=(x-a)Q+R}$ 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 ${\displaystyle f(x)=(x-a)Q}$.

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 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.