Section 2.3 Polynomial and Synthetic Division

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Section 2.3 Polynomial and Synthetic Division Powered By Docstoc
					Section 2.3 Polynomial and
    Synthetic Division
        What you should learn
• How to use long division to divide
  polynomials by other polynomials
• How to use synthetic division to divide
  polynomials by binomials of the form
                    (x – k)
• How to use the Remainder Theorem and the
  Factor Theorem
                        1. x goes into x3?    x2 times.
         x  2x  6
          2              2. Multiply (x-1) by x2.
                         3. Change sign, Add.
x 1 x  x  4x  6
       3  2
                        4. Bring down 4x.
     x x
   x  x
      3
      3   2
          2
                        5. x goes into 2x2? 2x times.
                        6. Multiply (x-1) by 2x.
    0  2 x  4x
           2
                         7. Change sign, Add

             
      2x  2x
           2
           2            8. Bring down -6.
                        9. x goes into 6x? 6 times.
        0  6x  6     10. Multiply (x-1) by 6.
                  
             6 x  6    11. Change sign, Add .

                   0
 Long Division.
                                  x 5
                      x  3 x  8 x  15
                              2

       Check
                                
                            x  3x
                              2


 ( x  3)( x  5)               5x  15
 x  5 x  3x  15
    2                              
                               5x  15
 x  8 x  15
   2                                  0
                          x 3x 9
                            2
 Divide.
             x  3 x  0 x  0 x  27
                     3      2


  x  27
   3                   
                   x  3x
                     3     2


    x3                  3x  0 x
                           2

                       3x  9 x
                           2
                              
                                9 x  27
x  3 x  27
       3
                                    
                              9x  27
                                      0
Long Division.
                               x 2
                    x  4 x  2x  8
                           2

   Check
                             
                        x  4x
                           2


 ( x  2)( x  4)             2x  8
 x  4x  2x  8
     2                        2x  8
                             2 x  8
 x  2x  8
   2                                0
Example
  p  2 p  20
    2
                                               44
                            =         p 4
     p6                                      p6
                           p  6 p  2 p  20
                                   2

            Check                 p 6p
                                p  6 p
                                   22

                            44 
( p  6)( p  4)  ( p  6)
                            p6
                                    4 p  20
                               
                                     
                                     4 p  24
 p  4 p  6 p  24  44
     2
                                           44
  p  2 p  20
        2
    p  2 p  20  p  4  44
     2


       p6                         p6
           f ( x)           r ( x)
                   q( x) 
          d ( x)            d ( x)
p  2 p  20   p  4 p  6 
                                     44
 2
                                        ( p  6)
                                    p6
  p  2 p  20   p  4 p  6  44
   2



         f ( x)  d ( x) q ( x)  r ( x)
         The Division Algorithm
If f(x) and d(x) are polynomials such that d(x) ≠ 0,
   and the degree of d(x) is less than or equal to the
   degree of f(x), there exists a unique polynomials
   q(x) and r(x) such that

          f ( x)  d ( x) q ( x)  r ( x)
Where r(x) = 0 or the degree of r(x) is less than
 the degree of d(x).
          Proper and Improper
            f ( x)           r ( x)
                    q( x) 
            d ( x)           d ( x)
• Since the degree of f(x) is more than or equal
  to d(x), the rational expression f(x)/d(x) is
  improper.
• Since the degree of r(x) is less than than d(x),
  the rational expression r(x)/d(x) is proper.
               Synthetic Division
          Divide x4 – 10x2 – 2x + 4 by x + 3
     -3        1    0     -10    -2     4
                   -3      +9     3     -3

               1   -3      -1     1      1

x  10 x  2 x  4
 4         2
                                       1
                    x  3x  x  1 
                      3    2

      x3                             x3
                             Long Division.




        1        -2   -8                     x 1
3
                 3     3          x  3 x  2x  8
                                         2


                                       x  3x
                                         2
                                           
         1   1        -5
                                               x 8
                                                 
                                              x 3
    f ( x)  x  2 x  8
             2
                                                 5
    f (3)  (3)  2(3)  8
                2


            9  6 8
            5
       The Remainder Theorem
If a polynomial f(x) is divided by x – k, the
   remainder is r = f(k).
                                            1x
f ( x)  x  2 x  8
           2                   x  3 x  2x  8
                                      2


f (3)  (3)  2(3)  8
            2                       x  3x
                                      2
                                        
        9  6 8                           x 8
        5                                   
                                           x 3
                                              5
          The Factor Theorem
A polynomial f(x) has a factor (x – k) if and only
  if f(k) = 0.
Show that (x – 2) and (x + 3) are factors of
          f(x) = 2x4 + 7x3 – 4x2 – 27x – 18


    +2      2     7     -4      -27 -18
                 4      22      36      18

            2    11     18       9       0
    Show that (x – 2) and (x + 3) are factors of
           f(x) = 2x4 + 7x3 – 4x2 – 27x – 18

+2                    2   7    -4   -27 -18
                          4    22   36     18

-3                    2   11   18   9
                          -6 -15    -9
                                          2 x 4  7 x3  4 x 2  27 x  18
                      2   5    3    0    (2 x 3  11 x 2  18 x  9)( x  2)
                                           (2 x 2  5 x  3)( x  2)( x  3)
Example 6 continued
                                          (2 x  3)(x  1)(x  2)(x  3)
  Uses of the Remainder in Synthetic
               Division
The remainder r, obtained in synthetic division
   of f(x) by (x – k), provides the following
   information.
1. r = f(k)
2. If r = 0 then (x – k) is a factor of f(x).
3. If r = 0 then (k, 0) is an x intercept of the
   graph of f.
        Fun with SYN and the TI-83
    f ( x)  x 2  8 x  15    f (3) 

•   Use SYN program to calculate f(-3)
•   [STAT] > Edit
•   Enter 1, 8, 15 into L1, then [2nd][QUIT]
•   Run SYN
•   Enter -3
     Fun with SYN and the TI-83
  f ( x)  15 x 4  10 x 3  6 x 2  14

• Use SYN program to calculate f(-2/3)
• [STAT] > Edit
• Enter 15, 10, -6, 0, 14 into L1, then
  [2nd][QUIT]
• Run SYN
• Enter 2/3
             2.3 Homework
• 1-67 odd

				
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posted:11/8/2012
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