Factor theorem

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 ) / ( x − a ) , for example using polynomial long division or synthetic division.
4. Conclude that any root x ≠ 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 at

To do this you 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:

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

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 + 7 x 2 + 8 x + 2.

The next two roots can be found by algebraically dividing x 3 + 7 x 2 + 8 x + 2 by ( x + 1 ) to get a quadratic, which can be solved directly, by the factor theorem or by the quadratic formula.

and therefore ( x + 1 ) and x 2 + 6 x + 2 are the factors of x 3 + 7 x 2 + 8 x + 2.