Polynomial Remainder Theorem

 

Motivation:

 

                        f(x) = x+3 . Let’s  consider the following situation.

 

                    If we set x = 1, then we have

                                         f(1) = 1 + 3 = 4

 

                    If  we set x=2, then we have

                                         f(2) = 2 + 3 = 5

 

 

                    If we divide f(x) by  (x-1),  we have

                             f(x) = x+3 = (x-1)  + 4

 

                    If we divide  f(x) by  (x-2),  we have

                             f(x) = x+3 = (x-2) + 5

 

                     Anything suspicious?  In general, we have the following theorem

 

 

 

Theorem ( Polynomial Remainder Theorem )

          Let f(x) be a polynomial of  x.  If  f(x) is divided by (x-a),  then the remainder polynomial is   f(a) .

 

       Proof:

                          When we divide f(x) by (x-a),  we know that the remainder polynomial is a constant because the degree of (x-a) is 1.  Thus, we have the following relationship

 

                                f(x) = (x-a) q(x) + C

                                         where q(x) is the quotient polynomial and C is a constant.

 

                      The equality should hold no matter what value we set for x.

                      Set x=a,  we have

 

                                f(a) = (a-a)q(x) + C = 0 + C = C

 

                      Thus,    f(a) = C .

 

 

                     And let’s check some simple application of this theorem.

 

 

 

 

 

 

Ex1.    (x¹⁰⁰ + 2) divided by (x-1)

 

                    The remainder polynomial is 

                             1¹⁰⁰ + 2 = 1 + 2 = 3

 

 

Ex2:    ( xⁿ -1 ) divided by (x-1)      ( n>1)

                           1¹⁰⁰ – 1 = 0

              Thus, we have the following result

 

                             xⁿ -1 = (x-1)q(x)

 

              Set x=10 and we have

                           10ⁿ -1 = 9 q(10)

 

              In other words,   9 is a divisor of  (10ⁿ -1 ).

 

 

Ex3:   8634357 divided by 3

                 Let  f(x) = 7 + 5x + 3x² + 4x³ + 3x⁴ +   6x⁵ + 8x⁶ .  

 

                Observe that

                       8634357 = f(10)

 

                And from polynomial remainder theorem, we have

                       f(x) = (x-1)q(x) + (8+6+3+4+3+5+7)

                              = (x-1)q(x) + 36

 

                Thus,

                              8634357 = f(10) = (10-1)q(10) + 36 = 9q(10) + 36

 

                 The sum of the coefficients of  f(x) can determine the result !

 

               So, the remainder is 0.

 

 

 

 

 

 

Ex4:  8634357 divided by 11

 

            The approach is similar to the previous example.

            Let f(x) = 7 + 5x + 3x² + 4x³ + 3x⁴ +   6x⁵ + 8x⁶ .

 

                         8634357 = f(10)

           And let’s divide f(x)  by (x+1).  We have

 

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

                           = (x+1) q(x) + ( 7-5+3-4+3-6+8 )

 

                   f(x) = (x+1) q(x) + (21-15) = (x+1)q(x) + 6

 

           Then

                       f(10) = (10+1)q(10) + 6 = 11q(10) + 6

 

            The remainder is 6.