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

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 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 $x-(-3-{\sqrt {7}}).$

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.

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[1] Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN 0-13-370149-2.

[2] Sehgal, V K; Gupta, Sonal, Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p. 119, ISBN 978-81-317-2816-1.

[3] Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p. 142, ISBN 81-7008-629-9.