Factor theorem

In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if $f(x)$ is a 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.{{citation|first=Michael|last=Sullivan|title=Algebra and Trigonometry|page=381|publisher=Prentice Hall|year=1996|isbn=0-13-370149-2}}{{citation |last1=Sehgal |first1=V K |title=Longman ICSE Mathematics Class 10 |page=119 |publisher=Dorling Kindersley (India) |isbn=978-81-317-2816-1 |last2=Gupta |first2=Sonal|date=September 2009 }}.

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:{{citation|first=R. K.|last=Bansal|title=Comprehensive Mathematics IX|page=142|publisher=Laxmi Publications|isbn=81-7008-629-9}}.

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

Proofs

Several proofs of the theorem are presented here.

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.

= Proof 1 =

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, $f(x) = f((x - a) + a) = (x - a)\cdot g(x - a)$ .

= Proof 2 =

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 $f(X)$ .

= Proof 3 =

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 $\deg(R) < \deg(x - a)$ . Since $\deg(R) < \deg(x - a)$ , 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.

= 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

Category:Theorems about polynomials