Theorems about Root of Polynomial

 

Definition ( Zero and Root )

             Let f(x) be a polynomial of x.   If  f(c) = 0,  then we say that c is a  zero of f(x) or root of the equation f(x)=0. In other words,  c is satisfying the following equation

 

                                               f(x) = 0 

 

              Sometimes, the term zero or root is used interchangeably.

 

 

Theorem ( Factor Theorem )

             Let f(x) be a polynomial of x.

                        (x-c) is a factor of f(x)  if and only if  f(c) = 0 .

 

Theorem      Let f(x) be a polynomial of x and c, d  Z,   c  0.  Then

                   The remainder polynomial of f(x) divided by (cx-d) is   f(  ) .

    Proof:

                         Via polynomial division, we can have

 

                             f(x) = (cx-d) q(x) + e  , where q(x) is a polynomial of x

 

                         Let set x=  :

                                  f(  ) = ( c  -d ) q(  ) + e = e

 

                        Thus, the remainder is  f(  ).

 

Theorem ( Rational Zero Theorem )

                 Let  

                          f(x)=  a_nxⁿ + a_n-1x^n-1 + ... + a₀      and f(x)  Z[x] ,  a_n   0 .

                   In other words, all the coefficients of f(x) are integers.

 

                  If  f(  ) = 0  and gcd( c, d ) = 1,  c  0,  then

                                     c | a_n  and d | a₀ .

 

         Proof:

                            f(  ) = 0

                       f(x) = (cx-d) q(x)

                                  where q(x) is a polynomial. But  how about the coefficients of q(x)?

 

                    There is no doubt that  coefficients of q(x) are rational numbers.

                    If people can be persuaded that the coefficients of q(x) are integers, the rest of the proof is easy.                          

 

                    Let’s consider the following things.  When we say two polynomials are equal, it means the coefficient of each term for both polynomials should be equal.  Furthermore,  when we set x to any value,  the results from the two equal polynomials must be the same.

 

                    Assume there exists an integer k such that

 

                                      q(k) =   ,  where s and t are integers, gcd( s, t) = 1, s  0

                    Set x=k :

                              f(k) = (ck-d ) q(k)

 

                   

                          Since the coefficients of f(x) are integers, then f(k) is an integer.

                          But on the right hand side, 

                                           c , d : integers

                                      (ck-d) is an integer

                          Furthermore,  q(k) =  and gcd(s,t)= 1

                             (ck-d)q(k)  is not an integer

                                But we have f(k) as an integer  ( contradiction )

 

                    Thus, the assumption does not hold.  In other words,  for every integer k,  q(k) is an integer.

 

                     So, we can conclude that the coefficients are integers.

 

 

                         We can represent q(x) as

 

                            q(x) = b_n-1x^n-1 + b_n-2x^n-2 + … + b₀

 

                       Then

                                  a_n = cb_n-1 ,  -db₀ = a₀

                             c | a_n and  d | a₀

 

 

 

Note:   The converse of the theorem is not true. But you can use this theorem to find the candidates of rational roots of a polynomial.

 

 

 

Theorem ( Intermediate Value Theorem )

                f(x)   R[x] .    If  f(a)f(b) < 0,  then there exists c between a and b such that

                                      f(c) = 0

 

      Proof:

                     The proof of this theorem will require some result from Calculus. In Calculus, it will show that polynomials are continuous functions. And every continuous function will have this “intermediate value theorem”.

 

                     You can think in this way: 

                                       f(a)f(b)<0

                     That means one of them is positive and one of them is negative.

                     For a continuous function, the transition from positive to negative or from negative to positive must pass through 0.

 

                    This theorem is helpful for identifying the location of the root(s) .

 

 

Theorem ( Fundamental Theorem of Algebra )

             Let f(x) be a non-constant polynomial in C[x].  Then there must exists c   C such that  f(c) = 0.

             In other words, for any non-constant polynomial f(x) in C[x]  with  order n,  there exist n roots in C for it ( repeated roots are counted ).

 

      Proof:

                      We just introduce the proof without going into details.  Gauss proved this theorem in his doctoral dissertation in the year 1799.   Nowadays, there are many tools to prove this theorem.

 

                      In Galois theory,  people can try to construct an extension field of C by finding contradiction with wrong assumption. That can show that the field C is good enough to let all the polynomials in C[x] have roots in C.

 

                      Another approach is from Complex analysis by using Liouville’s theorem. That can help to establish the result that a polynomial with no zero is constant.

 

 

 

 

 

 

 

 

 

 

 

Example:  Find the roots of  f(x)=2x³ -5x² –x+6

                   The divisors of 2 are   1,  2;

                   The divisors of 6 are  1,  2,  3,  6

                So, the candidates of roots :  1,  2,  3,  6,

               

x	-3	-2	-1	0	1	2	3
f(x)	-90	-28	0	6	2	0	12

 

                    f(x) = 2x³ -5x² –x +6

                          = (x+1)(2x² -7x+6)

                          = (x+1)(2x-3)(x-2)

 

                      Thus, the roots are  -1, 2,  .