Английская Википедия:Factor theorem
Шаблон:Short description In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if <math>f(x)</math> is a polynomial, then <math>x - a</math> is a factor of <math>f(x)</math> if and only if <math>f (a) = 0</math> (that is, <math>a</math> is a root of the polynomial). The theorem is a special case of the polynomial remainder theorem.[1][2]
The theorem results from basic properties of addition and multiplication. It follows that the theorem holds also when the coefficients and the element <math>a</math> 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 <math>f(X_1,\ldots,X_n)</math> and <math>g(X_2, \ldots,X_n)</math> are multivariate polynomials and <math>g</math> is independent of <math>X_1</math>, then <math>X_1 - g(X_2, \ldots,X_n)</math> is a factor of <math>f(X_1,\ldots,X_n)</math> if and only if <math>f(g(X_2, \ldots,X_n),X_2, \ldots,X_n)</math> is the zero polynomial.
Factorization of polynomials
Шаблон:Main 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]
- Deduce the candidate of zero <math>a</math> of the polynomial <math>f</math> from its leading coefficient <math>a_n</math> and constant term <math>a_0</math>. (See Rational Root Theorem.)
- Use the factor theorem to conclude that <math>(x-a)</math> is a factor of <math>f(x)</math>.
- Compute the polynomial <math display="inline"> g(x) = \dfrac{f(x)}{(x-a)} </math>, for example using polynomial long division or synthetic division.
- Conclude that any root <math>x \neq a</math> of <math>f(x)=0</math> is a root of <math>g(x)=0</math>. Since the polynomial degree of <math>g</math> is one less than that of <math>f</math>, it is "simpler" to find the remaining zeros by studying <math>g</math>.
Continuing the process until the polynomial <math>f</math> is factored completely, which all its factors is irreducible on <math>\mathbb{R}[x]</math> or <math>\mathbb{C}[x]</math>.
Example
Find the factors of <math>x^3 + 7x^2 + 8x + 2.</math>
Solution: Let <math>p(x)</math> be the above polynomial
- Constant term = 2
- Coefficient of <math>x^3=1 </math>
All possible factors of 2 are <math>\pm 1</math> and <math>\pm 2 </math>. Substituting <math>x=-1</math>, we get:
- <math>(-1)^3 + 7(-1)^2 + 8(-1) + 2 = 0</math>
So, <math>(x-(-1))</math>, i.e, <math>(x+1)</math> is a factor of <math>p(x)</math>. On dividing <math>p(x)</math> by <math>(x+1)</math>, we get
- Quotient = <math>x^2 + 6x + 2</math>
Hence, <math>p(x)=(x^2 + 6x + 2)(x+1)</math>
Out of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic <math>-3\pm \sqrt{7}.</math> Thus the three irreducible factors of the original polynomial are <math>x+1, </math> <math>x-(-3+\sqrt{7}),</math> and <math>x-(-3-\sqrt{7}).</math>
Proof
Several proofs of the theorem are presented here.
If <math>x-a</math> is a factor of <math>f(x), </math> it is immediate that <math>f(a)=0.</math> So, only the converse will be proved in the following.
Proof 1
This argument begins by verifying the theorem for <math>a = 0</math>. That is, it aims to show that for any polynomial <math>f(x)</math> for which <math>f(0) = 0</math> it is true that <math>f(x) =x\cdot g(x)</math> for some polynomial <math>g(x)</math>. To that end, write <math>f(x)</math> explicitly as <math>c_0 +c_1 x^1 + \dotsc + c_n x^n</math>. Now observe that <math>0 = f(0) = c_0</math>, so <math>c_0 = 0</math>. Thus, <math>f(x) = x(c_1 + c_2 x^1 + \dotsc + c_{n} x^{n-1}) = x \cdot g(x)</math>. This case is now proven.
What remains is to prove the theorem for general <math>a</math> by reducing to the <math>a = 0</math> case. To that end, observe that <math>f(x + a)</math> is a polynomial with a root at <math>x = 0</math>. By what has been shown above, it follows that <math>f(x + a) = x \cdot g(x)</math> for some polynomial <math>g(x)</math>. Finally, <math>f(x) = f((x - a) + a) = (x - a)\cdot g(x - a)</math>.
Proof 2
First, observe that whenever <math>x </math> and <math>y</math> belong to any commutative ring (the same one) then the identity <math>x^n - y^n = (x - y)(y^{n-1} + x^1 y^{n-2} + \dotsc + x^{n-2}y^{1} + x^{n-1})</math> is true. This is shown by multiplying out the brackets.
Let <math>f(X) \in R\left[ X \right]</math> where <math>R</math> is any commutative ring. Write <math>f(X) = \sum_i c_i X^i</math> for a sequence of coefficients <math>(c_i)_i</math>. Assume <math>f(a) = 0</math> for some <math>a \in R</math>. Observe then that <math>f(X) = f(X) - f(a) = \sum_{i} c_i(X^i - a^i)</math>. Observe that each summand has <math>X - a</math> as a factor by the factorisation of expressions of the form <math>x^n - y^n</math> that was discussed above. Thus, conclude that <math>X - a</math> is a factor of <math>f(X)</math>.
Proof 3
The theorem may be proved using Euclidean division of polynomials: Perform a Euclidean division of <math>f(x)</math> by <math>x - a</math> to obtain <math> f(x) = (x - a) Q + R</math> where <math>\deg(R) < \deg(x - a) </math>. Since <math>\deg(R) < \deg(x - a) </math>, it follows that <math>R</math> is constant. Finally, observe that <math>0 = f(a) = R</math>. So <math>f(x) = (x - a)Q</math>.
The Euclidean division above is possible in every commutative ring since <math>x - a</math> is a monic polynomial, and, therefore, the polynomial long division algorithm does not involves any division of coefficients.
Corollary of other theorems
It is also a corollary of the polynomial remainder theorem, but conversely can be used to show it.
When the polynomials are multivariate but the coefficients form an algebraically closed field, the Nullstellensatz is a significant and deep generalisation.
References
