Factor theorem

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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[edit]

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]

  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[edit]

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[edit]

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[edit]

  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 .