Synthetic Division and the Value of a Polynomial at a point

 

Motivation:

                      If f(x)=2x³-5x²-x+6 ,  what  is f(x-1)? Obviously, we will use (x-1)

                      to put into the original place for x:

 

                                 f(x-1)=2(x-1)³-5(x-1)²-(x-1)+6

                                          = 2x³-11x²+15x

 

                       If we define g(x)=f(x-1),  then g(x)=2x³-11x²+15x .

 

                       Let’s forget the previous exercise and focus on the following problem.

                       If  two polynomials g(x) and f(x) are with the relationship

                                       

                                              g(x)=f(x-1) and g(x)=2x³-11x²+15x  ,

 

                       Can you find f(x) ?

 

                              You can start by dividing g(x) by (x-1), the remainder will be the constant term of f(x); then you divide the quotient by (x-1) to get another remainder for the coefficient of x term of f(x) …

 

                             It all can be done by using “long division” for polynomials. However, we are going to introduce another method for division: “synthetic division”.

 

Example ( Synthetic Division )

                      g(x) =  2x³-11x²+15x . Find the quotient polynomial and remainder while g(x) is divided by (x-1) .

 

                                  2- 11+ 15+ 0  | 1         ---  divided by (x-1),  put 1 here                   

                                     +2  -  9 + 6

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

2-     9  + 6  |+ 6            --- remainder after divided by (x-1)

 

                     The quotient polynomial is  2x² – 9x + 6;

                     the remainder is 6.

 

                     The operation principle of synthetic division is:  when you  divide A(x) by B(x),  you just divide each other term in B(x) by  the leading coefficient of B(x) and put minus sign in the result as a multiplier.  Then use a series “multiplication” and “addition” to perform the work for “division”  ( In “long division”, we use “subtraction”. )

 

 

 

 

 

 

Example:                Let       g(x) =  2x³-11x²+15x  and  g(x)=f(x-1) . Find f(x).

 

                                  2- 11+ 15+ 0  | 1         ---  divided by (x-1),  put 1 here                   

                                     +2  -  9 + 6

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

3-     9  + 6  |+ 6            --- remainder after divided by (x-1)

                                      +2  - 7

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

                                  2 – 7  |-  1                   --- dividing the quotient by (x-1)

                                     + 2

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

                                  2  |-5

 

                            So,  g(x) = (3x² – 9x + 6)(x-1) + 6

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

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

                                           =  (2(x-1) -5)(x-1)² –(x-1) + 6

                                           = 2(x-1)³ – 5(x-1)² – (x-1) + 6

 

 

                        Thus,   f(x) = 2x³ – 5x² –x + 6 .

 

 

                        What is the benefit of doing that?  We see the following example.

 

Example:  Let       g(x) =  2x³-11x²+15x  . 

                  Evaluate  g(1.0002) down to 5 digits after decimal point.

 

           Sol:

                         From the result of the previous example:

                             g(x) = 2(x-1)³ – 5(x-1)² – (x-1) + 6

 

                          So,

                             g(1.0002) = 2  (0.0002)³ -5(0.0002)² – 0.0002 + 6

                                              5.99980