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.^[1]

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).^[2]

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:^[3]

"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.)
Use the factor theorem to conclude that $(x-a)$ is a factor of $f(x)$ .
Compute the polynomial $g(x) = f(x) \big/ (x-a)$ , 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 $g$ .

Example

Find the factors at

x^3 + 7x^2 + 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:

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, which can be solved directly, by the factor theorem or by the quadratic formula.

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

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

Formal version

Let $f$ be a one-variable polynomial with coefficients in a commutative ring $R$ , and let $a \in R$ . Then $f(a) = 0$ if and only if $f(x)=(x-a)g(x)$ for some polynomial $g$ . In this case, $g$ is determined uniquely.

As for the problem of algorithmically finding all roots, if $f$ is given and $a$ is known, then $g$ can be computed by polynomial long division; then one can compute the remaining roots of $f$ , including repeated roots, by factoring $g$ .

References

↑ Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN 0-13-370149-2 .
↑ Sehgal, V K; Gupta, Sonal, Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p. 119, ISBN 978-81-317-2816-1 .
↑ Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p. 142, ISBN 81-7008-629-9 .