Polynomial – Factorization

 

Motivation: 

                        Factorization of polynomial is to represent a polynomial into the product of several lower order polynomials.  Why do we want to do that? One of the reasons is that it is easier to understand some properties of the higher order polynomials by observing those lower order polynomials.  Let’s check the following question.

 

                       If  f(x)= x²-3x+2=(x-1)(x-2) <0 ,  what is the range of x to satisfy this inequality ?

 

                       We know that  f(x)=0  when x=1 or x=2.   So, it is natural that we will check if x satisfies the requirement by trying the following ranges:

 

                                      x < 1

                                   1< x < 2

                                        x > 2

 

                       

 

                          And we find out that    the range 1< x < 2 can satisfy the inequality.

                    But if we write f(x) as  (x² -3x +2 ) without factoring it,  it would not be easy to find out the answer.

 

Definition ( Factor )

                 If  a polynomial f(x) is divided by g(x) with remainder polynomial 0, then g(x) is a factor of f(x).  We denote it as  g(x)|f(x)  .

 

                 For a polynomial, if there exists no factor with order greater or equal to 1, we say that this polynomial is irreducible.  However, it is necessary to specify the “scope” when we say a polynomial is irreducible.  At this moment, we can think that it is due to the requirement on the coefficients the factor polynomials.   We use an example to explain that.

 

                       f(x) = x² -2

 

              If we require that the coefficients of factor polynomials are Integers ( denoted as Z ), then f(x) is irreducible.  To be complete, we say that it is irreducible in Z[x].  But let’s check

 

                   (x+  )(x-  ) = x² -2

                   f(x) is not irreducible in R[x]  ( R stands for the set of real numbers ) .

 

           In more advanced math courses,  it will be proved that  if f(x)  Z[x] is irreducible in Z[x], then it is also irreducible in Q[x] ( rational numbers ).

 

 

 

Definition ( Irreducible Polynomial )

                If a polynomial f(x) is said to be irreducible in Z[x] , it means there exists no non-constant polynomial with coefficients in integers such that it is a factor polynomial of f(x).

 

                Usually, if it is not mentioned specifically, we just focus on the factorization of polynomials by finding the factors with coefficients in integers.

 

 

 

 

Theorem ( Factor Theorem)

                Let f(x) be a polynomial of x.  If  f(c)=0,  then  (x-c)| f(x).

           In other words,  (x-c) is a factor of f(x).

 

         Proof:

                       It is just a trivial result from “Remainder Theorem of Polynomial”:

 

                                    f(x) = (x-c)q(x) + f(c)  = (x-c)q(x)

 

                       So,   (x-c)|f(x)

 

Example:   f(x)=x³-1 .  Factorize f(x) .

                

                  f(1)=0. Then (x-1) is a factor of f(x).

 

                  To factor f(x), we can divide f(x) by (x-1) via “long division”.

                   So,

                               f(x) = (x-1)(x² + x + 1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Factoring 2^nd order polynomials

 

                 Previously, we have    (ax+b)(cx+d) = (ac)x² + (ad+bc)x + bd   .

          We can try to factor a 2^nd order polynomial with the following diagram

 

                                  

 

                  And remember:

1.      using ac to check the coefficient of the term x²

2.      using  bd to check the coefficient of the constant term

3.      using the result of  “cross multiplication” ( ad+bc) to check the coefficient of x

 

                        If  all of them match, then  this second order polynomial can be factored into

                                    (ax+b)(cx+d)

 

Example:    Factorize   15x² – 19x + 6 .

 

                          It seems the following arrangement matches !

                                         3             -2

                                                

5                          -3

                                    --------------------------

 

                     Thus,  15x² – 19x + 6 = (3x-2)(5x-3)