Factor theorem

In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if f(x) is a (univariate) polynomial, then x - a is a factor of f(x) if and only if f (a) = 0 (that is, a is a root of the polynomial). The theorem is a special case of the polynomial remainder theorem. The theorem results from basic properties of addition and multiplication. It follows that the theorem holds also when the coefficients and the element a belong to any commutative ring, and not just a field. In particular, since multivariate polynomials can be viewed as univariate in one of their variables, the following generalization holds : If f(X_1,\ldots,X_n) and g(X_2, \ldots,X_n) are multivariate polynomials and g is independent of X_1 , then X_1 - g(X_2, \ldots,X_n) is a factor of f(X_1,\ldots,X_n) if and only if f(g(X_2, \ldots,X_n),X_2, \ldots,X_n) is the zero polynomial.

In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if f(x) is a (univariate) polynomial, then x - a is a factor of f(x) if and only if f (a) = 0 (that is, a is a root of the polynomial). The theorem is a special case of the polynomial remainder theorem. The theorem results from basic properties of addition and multiplication. It follows that the theorem holds also when the coefficients and the element a belong to any commutative ring, and not just a field. In particular, since multivariate polynomials can be viewed as univariate in one of their variables, the following generalization holds : If f(X_1,\ldots,X_n) and g(X_2, \ldots,X_n) are multivariate polynomials and g is independent of X_1 , then X_1 - g(X_2, \ldots,X_n) is a factor of f(X_1,\ldots,X_n) if and only if f(g(X_2, \ldots,X_n),X_2, \ldots,X_n) is the zero polynomial. ==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: Deduce the candidate of zero a of the polynomial f from its leading coefficient a_n and constant term a_0 . (See Rational root theorem.) Use the factor theorem to conclude that (x-a) is a factor of f(x) . Compute the polynomial g(x) = \dfrac{f(x)}{(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 . Continuing the process until the polynomial f is factored completely, which all its factors is irreducible on \mathbb{R}[x] or \mathbb{C}[x] . ===Example=== Find the factors of x^3 + 7x^2 + 8x + 2. Solution: Let p(x) be the above polynomial Constant term = 2 Coefficient of x^3=1 All possible factors of 2 are \pm 1 and \pm 2 . Substituting x=-1 , we get: (-1)^3 + 7(-1)^2 + 8(-1) + 2 = 0 So, (x-(-1)) , i.e., (x+1) is a factor of p(x) . On dividing p(x) by (x+1) , we get Quotient = x^2 + 6x + 2 Hence, p(x)=(x^2 + 6x + 2)(x+1) Out 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}). If x-a is a factor of f(x), it is immediate that f(a)=0. So, only the converse will be proved in the following. This proof begins by verifying the statement for a = 0 . That is, it will show that for any polynomial f(x) for which f(0) = 0 , there exists a polynomial g(x) such that f(x) =x\cdot g(x) . To that end, write f(x) explicitly as c_0 +c_1 x^1 + \dotsc + c_n x^n . Now observe that 0 = f(0) = c_0 , so c_0 = 0 . Thus, f(x) = x(c_1 + c_2 x^1 + \dotsc + c_{n} x^{n-1}) = x \cdot g(x) . This case is now proven. What remains is to prove the theorem for general a by reducing to the a = 0 case. To that end, observe that f(x + a) is a polynomial with a root at x = 0 . By what has been shown above, it follows that f(x + a) = x \cdot g(x) for some polynomial g(x) . Finally, f(x) = f((x - a) + a) = (x - a)\cdot g(x - a) . First, observe that whenever x and y belong to any commutative ring (the same one) then the identity x^n - y^n = (x - y)(y^{n-1} + x^1 y^{n-2} + \dotsc + x^{n-2}y^{1} + x^{n-1}) is true. This is shown by multiplying out the brackets. Let f(X) \in R\left[ X \right] where R is any commutative ring. Write f(X) = \sum_i c_i X^i for a sequence of coefficients (c_i)_i . Assume f(a) = 0 for some a \in R . Observe then that f(X) = f(X) - f(a) = \sum_{i} c_i(X^i - a^i) . Observe that each summand has X - a as a factor by the factorisation of expressions of the form x^n - y^n that was discussed above. Thus, conclude that X - a is a factor of f(X) . The theorem may be proved using Euclidean division of polynomials: Perform a Euclidean division of f(x) by (x-a) to obtain f(x) = (x - a) Q(x)+ R(x) where \deg(R) . Since \deg(R) , it follows that R is constant. Finally, observe that 0 = f(a) = R . So f(x) = (x - a)Q(x) . The Euclidean division above is possible in every commutative ring since (x - a) is a monic polynomial, and, therefore, the polynomial long division algorithm does not involve any division of coefficients. ==References==

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