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.

Factorization of polynomials

Main article: 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 math 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.
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

You wish to find the factors at

x³ + 7x² + 8x + 2.

To do this you would use trial and error 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:

x³ + 7x² + 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 x³ + 7x² + 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 x³ + 7x² + 8x + 2.

The next two roots can be found by algebraically dividing x³ + 7x² + 8x + 2 by (x + 1) to get a quadratic, which can be solved directly, by the factor theorem or by the quadratic equation.

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

and therefore (x + 1) and x² + 6x + 2 are the factors of x³ + 7x² + 8x + 2.

Formal version

Let f be a polynomial with complex coefficients, and a be in an integral domain (e.g. $a \in \mathbb{C}$). Then f(a) = 0 if and only if f(x) can be written in the form f(x) = (x − a)g(x) where g(x) is also a polynomial. g is determined uniquely.

This indicates that those a for which f(a) = 0 are precisely the roots of f(x). Repeated roots can be found by application of the theorem to the quotient g, which may be found by polynomial long division.