# Factor theorem

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.

The factor theorem states that a polynomial $f(x)$ has a factor $(x-\alpha )$ if and only if $f(\alpha )=0$ (i.e. $\alpha$ is a root).

More generally, a bivariate polynomial $g(x,y)$ has a factor $x-y$ if and only $g(x,x)$ is the zero polynomial. The above theorem is the case where $g(x,y)=f(x)-f(y).$ ## 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 its all 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 $x-(-3-{\sqrt {7}}).$ 