Factor theorem

☆ Save On Wikipedia ↗

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

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:[3]

  1. Deduce the candidate of zero a {\displaystyle a} {\displaystyle a} of the polynomial f {\displaystyle f} {\displaystyle f} from its leading coefficient a n {\displaystyle a_{n}} {\displaystyle a_{n}} and constant term a 0 {\displaystyle a_{0}} {\displaystyle a_{0}}. (See Rational root theorem.)
  2. Use the factor theorem to conclude that ( x − a ) {\displaystyle (x-a)} {\displaystyle (x-a)} is a factor of f ( x ) {\displaystyle f(x)} {\displaystyle f(x)}.
  3. Compute the polynomial g ( x ) = f ( x ) ( x − a ) {\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 x ≠ a {\displaystyle x\neq a} {\displaystyle x\neq a} of f ( x ) = 0 {\displaystyle f(x)=0} {\displaystyle f(x)=0} is a root of g ( x ) = 0 {\displaystyle g(x)=0} {\displaystyle g(x)=0}. Since the polynomial degree of g {\displaystyle g} {\displaystyle g} is one less than that of f {\displaystyle f} {\displaystyle f}, it is "simpler" to find the remaining zeros by studying g {\displaystyle g} {\displaystyle g}.

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

Example

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

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

Constant term = 2
Coefficient of x 3 = 1 {\displaystyle x^{3}=1} {\displaystyle x^{3}=1}

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

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

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

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

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

Proofs

Several proofs of the theorem are presented here.

If x − a {\displaystyle x-a} {\displaystyle x-a} is a factor of f ( x ) , {\displaystyle f(x),} {\displaystyle f(x),} it is immediate that f ( a ) = 0. {\displaystyle f(a)=0.} {\displaystyle 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 {\displaystyle a=0} {\displaystyle a=0}. That is, it will show that for any polynomial f ( x ) {\displaystyle f(x)} {\displaystyle f(x)} for which f ( 0 ) = 0 {\displaystyle f(0)=0} {\displaystyle f(0)=0}, there exists a polynomial g ( x ) {\displaystyle g(x)} {\displaystyle g(x)} such that f ( x ) = x ⋅ g ( x ) {\displaystyle f(x)=x\cdot g(x)} {\displaystyle f(x)=x\cdot g(x)}. To that end, write f ( x ) {\displaystyle f(x)} {\displaystyle f(x)} explicitly as c 0 + c 1 x 1 + … + c n x n {\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 0 = f ( 0 ) = c 0 {\displaystyle 0=f(0)=c_{0}} {\displaystyle 0=f(0)=c_{0}}, so c 0 = 0 {\displaystyle c_{0}=0} {\displaystyle c_{0}=0}. Thus, f ( x ) = x ( c 1 + c 2 x 1 + … + c n x n − 1 ) = x ⋅ g ( x ) {\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 a {\displaystyle a} {\displaystyle a} by reducing to the a = 0 {\displaystyle a=0} {\displaystyle a=0} case. To that end, observe that f ( x + a ) {\displaystyle f(x+a)} {\displaystyle f(x+a)} is a polynomial with a root at x = 0 {\displaystyle x=0} {\displaystyle x=0}. By what has been shown above, it follows that f ( x + a ) = x ⋅ g ( x ) {\displaystyle f(x+a)=x\cdot g(x)} {\displaystyle f(x+a)=x\cdot g(x)} for some polynomial g ( x ) {\displaystyle g(x)} {\displaystyle g(x)}. Finally, f ( x ) = f ( ( x − a ) + a ) = ( x − a ) ⋅ g ( x − a ) {\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)}.

Proof 2

First, observe that whenever x {\displaystyle x} {\displaystyle x} and y {\displaystyle y} {\displaystyle 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 + … + 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})} {\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 f ( X ) ∈ R [ X ] {\displaystyle f(X)\in R\left[X\right]} {\displaystyle f(X)\in R\left[X\right]} where R {\displaystyle R} {\displaystyle R} is any commutative ring. Write f ( X ) = ∑ i c i X i {\displaystyle f(X)=\sum _{i}c_{i}X^{i}} {\displaystyle f(X)=\sum _{i}c_{i}X^{i}} for a sequence of coefficients ( c i ) i {\displaystyle (c_{i})_{i}} {\displaystyle (c_{i})_{i}}. Assume f ( a ) = 0 {\displaystyle f(a)=0} {\displaystyle f(a)=0} for some a ∈ R {\displaystyle a\in R} {\displaystyle a\in R}. Observe then that f ( X ) = f ( X ) − f ( a ) = ∑ i c i ( X i − a i ) {\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 X − a {\displaystyle X-a} {\displaystyle X-a} as a factor by the factorisation of expressions of the form x n − y n {\displaystyle x^{n}-y^{n}} {\displaystyle x^{n}-y^{n}} that was discussed above. Thus, conclude that X − a {\displaystyle X-a} {\displaystyle X-a} is a factor of f ( X ) {\displaystyle f(X)} {\displaystyle f(X)}.

Proof 3

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

  1. Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN 0-13-370149-2
  2. Sehgal, V K; Gupta, Sonal (September 2009), 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.