# Factor theorem

In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if $f(x)$ is a 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 $f(X_{1},\ldots ,X_{n})$ 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 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)=f(x)/(x−a) {\textstyle g(x)={\dfrac {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 $\mathbb {R} [x]$ 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 $x-(-3-{\sqrt {7}}).$

## 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 $c_{0}+c_{1}x^{1}+\dotsc +c_{n}x^{n}$. 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, $f(x)=f((x-a)+a)=(x-a)\cdot g(x-a)$.

### 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 $f(X)$.

### 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.