# 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 $(x-k)$ if and only if $f(k)=0$ (i.e. $k$ is a root).

## 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. "Guess" a zero $a$  of the polynomial $f$ . (In general, this can be very hard, but maths textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)
2. Use the factor theorem to conclude that $(x-a)$  is a factor of $f(x)$ .
3. Compute the polynomial $g(x)=f(x){\big /}(x-a)$ , 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 $g$ .

### Example

Find the factors of

$x^{3}+7x^{2}+8x+2.$

To do this one would use trial and error (or the rational root theorem) to find the first x value that causes the expression to equal zero. To find out if $(x-1)$  is a factor, substitute $x=1$  into the polynomial above:

$x^{3}+7x^{2}+8x+2=(1)^{3}+7(1)^{2}+8(1)+2$
$=1+7+8+2$
$=18.$

As this is equal to 18 and not 0. This means $(x-1)$  is not a factor of $x^{3}+7x^{2}+8x+2$ . So, we next try $(x+1)$  (substituting $x=-1$  into the polynomial):

$(-1)^{3}+7(-1)^{2}+8(-1)+2.$

This is equal to $0$ . Therefore $x-(-1)$ , which is to say $x+1$ , is a factor, and $-1$  is a root of $x^{3}+7x^{2}+8x+2.$

The next two roots can be found by algebraically dividing $x^{3}+7x^{2}+8x+2$  by $(x+1)$  to get a quadratic:

${x^{3}+7x^{2}+8x+2 \over x+1}=x^{2}+6x+2,$

and therefore $(x+1)$  and $x^{2}+6x+2$  are factors of $x^{3}+7x^{2}+8x+2.$  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 $x-(-3-{\sqrt {7}}).$