Factor theorem

In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if ${\displaystyle f(x)}${\displaystyle f(x)} is a polynomial, then ${\displaystyle x-a}${\displaystyle x-a} is a factor of ${\displaystyle f(x)}${\displaystyle f(x)} if and only if ${\displaystyle f(a)=0}${\displaystyle f(a)=0} (that is, ${\displaystyle a}${\displaystyle a} 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 ${\displaystyle a}${\displaystyle 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 ${\displaystyle f(X_{1},\ldots ,X_{n})}${\displaystyle f(X_{1},\ldots ,X_{n})} and ${\displaystyle g(X_{2},\ldots ,X_{n})}${\displaystyle g(X_{2},\ldots ,X_{n})} are multivariate polynomials and ${\displaystyle g}${\displaystyle g} is independent of ${\displaystyle X_{1}}${\displaystyle X_{1}}, then ${\displaystyle X_{1}-g(X_{2},\ldots ,X_{n})}${\displaystyle X_{1}-g(X_{2},\ldots ,X_{n})} is a factor of ${\displaystyle f(X_{1},\ldots ,X_{n})}${\displaystyle f(X_{1},\ldots ,X_{n})} if and only if ${\displaystyle f(g(X_{2},\ldots ,X_{n}),X_{2},\ldots ,X_{n})}${\displaystyle f(g(X_{2},\ldots ,X_{n}),X_{2},\ldots ,X_{n})} is the zero polynomial.

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. Deduce the candidate of zero ${\displaystyle a}${\displaystyle a} of the polynomial ${\displaystyle f}${\displaystyle f} from its leading coefficient ${\displaystyle a_{n}}${\displaystyle a_{n}} and constant term ${\displaystyle a_{0}}${\displaystyle a_{0}}. (See Rational Root Theorem.)
2. Use the factor theorem to conclude that ${\displaystyle (x-a)}${\displaystyle (x-a)} is a factor of ${\displaystyle f(x)}${\displaystyle f(x)}.
3. Compute the polynomial ${\textstyle g(x)={\dfrac {f(x)}{(x-a)}}}${\textstyle g(x)={\dfrac {f(x)}{(x-a)}}}, for example using polynomial long division or synthetic division.
4. Conclude that any root ${\displaystyle x\neq a}${\displaystyle x\neq a} of ${\displaystyle f(x)=0}${\displaystyle f(x)=0} is a root of ${\displaystyle g(x)=0}${\displaystyle g(x)=0}. Since the polynomial degree of ${\displaystyle g}${\displaystyle g} is one less than that of ${\displaystyle f}${\displaystyle f}, it is "simpler" to find the remaining zeros by studying ${\displaystyle g}${\displaystyle g}.

Continuing the process until the polynomial ${\displaystyle f}${\displaystyle f} is factored completely, which all its factors is irreducible on ${\displaystyle \mathbb {R} [x]}${\displaystyle \mathbb {R} [x]} or ${\displaystyle \mathbb {C} [x]}${\displaystyle \mathbb {C} [x]}.

Find the factors of ${\displaystyle x^{3}+7x^{2}+8x+2.}${\displaystyle x^{3}+7x^{2}+8x+2.}

Solution: Let ${\displaystyle p(x)}${\displaystyle p(x)} be the above polynomial

Constant term = 2

Coefficient of ${\displaystyle x^{3}=1}${\displaystyle x^{3}=1}

All possible factors of 2 are ${\displaystyle \pm 1}${\displaystyle \pm 1} and ${\displaystyle \pm 2}${\displaystyle \pm 2}. Substituting ${\displaystyle x=-1}${\displaystyle x=-1}, we get:

${\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2=0}${\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2=0}

So, ${\displaystyle (x-(-1))}${\displaystyle (x-(-1))}, i.e, ${\displaystyle (x+1)}${\displaystyle (x+1)} is a factor of ${\displaystyle p(x)}${\displaystyle p(x)}. On dividing ${\displaystyle p(x)}${\displaystyle p(x)} by ${\displaystyle (x+1)}${\displaystyle (x+1)}, we get

Quotient = ${\displaystyle x^{2}+6x+2}${\displaystyle x^{2}+6x+2}

Hence, ${\displaystyle p(x)=(x^{2}+6x+2)(x+1)}${\displaystyle 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 ${\displaystyle -3\pm {\sqrt {7}}.}${\displaystyle -3\pm {\sqrt {7}}.} Thus the three irreducible factors of the original polynomial are ${\displaystyle x+1,}${\displaystyle x+1,} ${\displaystyle x-(-3+{\sqrt {7}}),}${\displaystyle x-(-3+{\sqrt {7}}),} and ${\displaystyle x-(-3-{\sqrt {7}}).}${\displaystyle x-(-3-{\sqrt {7}}).}

Several proofs of the theorem are presented here.

If ${\displaystyle x-a}${\displaystyle x-a} is a factor of ${\displaystyle f(x),}${\displaystyle f(x),} it is immediate that ${\displaystyle f(a)=0.}${\displaystyle f(a)=0.} So, only the converse will be proved in the following.

This proof begins by verifying the statement for ${\displaystyle a=0}${\displaystyle a=0}. That is, it will show that for any polynomial ${\displaystyle f(x)}${\displaystyle f(x)} for which ${\displaystyle f(0)=0}${\displaystyle f(0)=0}, there exists a polynomial ${\displaystyle g(x)}${\displaystyle g(x)} such that ${\displaystyle f(x)=x\cdot g(x)}${\displaystyle f(x)=x\cdot g(x)}. To that end, write ${\displaystyle f(x)}${\displaystyle f(x)} explicitly as ${\displaystyle c_{0}+c_{1}x^{1}+\dotsc +c_{n}x^{n}}${\displaystyle c_{0}+c_{1}x^{1}+\dotsc +c_{n}x^{n}}. Now observe that ${\displaystyle 0=f(0)=c_{0}}${\displaystyle 0=f(0)=c_{0}}, so ${\displaystyle c_{0}=0}${\displaystyle c_{0}=0}. Thus, ${\displaystyle f(x)=x(c_{1}+c_{2}x^{1}+\dotsc +c_{n}x^{n-1})=x\cdot g(x)}${\displaystyle 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 ${\displaystyle a}${\displaystyle a} by reducing to the ${\displaystyle a=0}${\displaystyle a=0} case. To that end, observe that ${\displaystyle f(x+a)}${\displaystyle f(x+a)} is a polynomial with a root at ${\displaystyle x=0}${\displaystyle x=0}. By what has been shown above, it follows that ${\displaystyle f(x+a)=x\cdot g(x)}${\displaystyle f(x+a)=x\cdot g(x)} for some polynomial ${\displaystyle g(x)}${\displaystyle g(x)}. Finally, ${\displaystyle f(x)=f((x-a)+a)=(x-a)\cdot g(x-a)}${\displaystyle f(x)=f((x-a)+a)=(x-a)\cdot g(x-a)}.

First, observe that whenever ${\displaystyle x}${\displaystyle x} and ${\displaystyle y}${\displaystyle y} belong to any commutative ring (the same one) then the identity ${\displaystyle x^{n}-y^{n}=(x-y)(y^{n-1}+x^{1}y^{n-2}+\dotsc +x^{n-2}y^{1}+x^{n-1})}${\displaystyle 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 ${\displaystyle f(X)\in R\left[X\right]}${\displaystyle f(X)\in R\left[X\right]} where ${\displaystyle R}${\displaystyle R} is any commutative ring. Write ${\displaystyle f(X)=\sum _{i}c_{i}X^{i}}${\displaystyle f(X)=\sum _{i}c_{i}X^{i}} for a sequence of coefficients ${\displaystyle (c_{i})_{i}}${\displaystyle (c_{i})_{i}}. Assume ${\displaystyle f(a)=0}${\displaystyle f(a)=0} for some ${\displaystyle a\in R}${\displaystyle a\in R}. Observe then that ${\displaystyle f(X)=f(X)-f(a)=\sum _{i}c_{i}(X^{i}-a^{i})}${\displaystyle f(X)=f(X)-f(a)=\sum _{i}c_{i}(X^{i}-a^{i})}. Observe that each summand has ${\displaystyle X-a}${\displaystyle X-a} as a factor by the factorisation of expressions of the form ${\displaystyle x^{n}-y^{n}}${\displaystyle x^{n}-y^{n}} that was discussed above. Thus, conclude that ${\displaystyle X-a}${\displaystyle X-a} is a factor of ${\displaystyle f(X)}${\displaystyle f(X)}.

The theorem may be proved using Euclidean division of polynomials: Perform a Euclidean division of ${\displaystyle f(x)}${\displaystyle f(x)} by ${\displaystyle (x-a)}${\displaystyle (x-a)} to obtain ${\displaystyle f(x)=(x-a)Q(x)+R(x)}${\displaystyle f(x)=(x-a)Q(x)+R(x)} where ${\displaystyle \deg(R)<\deg(x-a)}${\displaystyle \deg(R)<\deg(x-a)}. Since ${\displaystyle \deg(R)<\deg(x-a)}${\displaystyle \deg(R)<\deg(x-a)}, it follows that ${\displaystyle R}${\displaystyle R} is constant. Finally, observe that ${\displaystyle 0=f(a)=R}${\displaystyle 0=f(a)=R}. So ${\displaystyle f(x)=(x-a)Q(x)}${\displaystyle f(x)=(x-a)Q(x)}.

The Euclidean division above is possible in every commutative ring since ${\displaystyle (x-a)}${\displaystyle (x-a)} is a monic polynomial, and, therefore, the polynomial long division algorithm does not involve any division of coefficients.

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.

• Theorems about polynomials

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