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					          Finding Zeros of Polynomials




 Jeff Bivin -- LZHS
Last updated: 12-4-07
 Divide: 4 x  5 x  9 x  18  x  2 x  4
                         4    2                  2


                             4 x  8x  5
                                  2


           x  2 x  4 4 x  0 x  5 x  9 x  18
                     2            4   3      2

                    - ( 4 x 4  8 x 3  16 x 2 )
                                 8 x  11 x  9 x
                                      3          2

                          - (  8 x  16 x  32 x )
                                     3          2


                                            5 x  23 x  18
                                                2

           2             13 x  2
        4 x  8x  5  x2  2 x  4     - ( 5 x 2  10 x  20 )
                                                     13x  2
Jeff Bivin -- LZHS
                                      Divide:
 6 x  7 x  18 x  25 x  12 x  18  2 x  x  3
        5             4       3            2            2



                           3x  5 x  2 x  6
                             3         2




                     6 x  13 x  4 x  1  3x  1
                       3          2




                           2 x  5x  3 
                             2                    2
                                               3 x 1



Jeff Bivin -- LZHS
     Use Synthetic Division to Divide
                     x  4 x  5x  3  x  2
                      3        2

 x+2=0
                                        1   4      -5      3
      x = -2
                          -2                -2     -4     18
                                        1   2      -9     21
                                                        remainder

                                   x  2x  9 
                                    2              21
                                                  x2



Jeff Bivin -- LZHS
     Use Synthetic Division to Divide
             2 x  11 x  3x  36 by x  3
                     3   2


   x-3=0
                               2    -11        3    36
         x=3
                         3            6       -15   -36
                               2     -5       -12    0
     2 x  5 x  12
            2


    Factored Form:           x  3 2 x  5x  12 
                                          2


                              x  32 x  3x  4
Jeff Bivin -- LZHS
                  Factor
x  4 x  15 x  18 if x  1 is a factor
 3     2

 x+1=0
      x = -1               1       4    -15   -18
                     -1            -1    -3   18
                           1       3    -18    0

                      x  1x  3x  18 
                               2


                      x  1x  3x  6
Jeff Bivin -- LZHS
                           Find the zeros of
                     f ( x)  6 x  7 x  43 x  30
                                 3     2

                         if x  3 is one zero
 x=3
  x–3=0
                                6        -7   -43   30
                          3          18       33    -30
                                6    11       -10     0

                          x  3 6 x  11x  10 
                                     2


                           x  32 x  53x  2
Jeff Bivin -- LZHS
                        Factor Theorem

                     A polynomial f(x) has a factor x – k
                               if and only if
                                  f(k) = 0.




Jeff Bivin -- LZHS
                      Rational Zero Theorem
                     If f(x) = anxn + . . . + a1x + a0 has
                     integer coefficients, then every
                     rational zero of f(x) has the
                     following form:
                        p     factor of constant term a0
                          =
                        q   factor of leading coefficient an




Jeff Bivin -- LZHS
        List the possible rational zeros
                     f ( x)  6 x  5 x  9 x  4 x  15
                                 4      3      2



                      15
                       p     factor of1, 3, 5, 15 term a0
                                       constant
                         =
                      6q               1, 2, coefficient an
                           factor of leading3, 6

        1  1        33   5
                            5 15  15   1    3   5    15
         1            1     1       1    2     2    2      2
         1             3     5    15      1    3     5    15
                                                
         3             3     3     3      6    6     6     6

Jeff Bivin -- LZHS
        List the possible rational zeros
                 f ( x)  9 x  5 x  9 x  4 x  24
                                 12        9      4         3



                       24
                        p        1, 2, 3, 4, 6, 8, term
                              factor of constant12, 24a0
                          =
                       9q   factor of leading coefficient an
                                          1, 3, 9

       1  2              3 4       6  8  12  24

         1             2     4       8      1     2     4         8
                                                         
         3             3     3       3      9     9     9         9

Jeff Bivin -- LZHS
                         Find all real zeros
                     f ( x)  8 x  2 x  21 x  7 x  3
                                4         3           2



                       p
                       3                  1, 3
                             factor of constant term a0
                         =
                       q
                       8   factor of leading coefficient an
                                       1, 2, 4, 8
                          1     3     1     3     1         3
       1  3                                        
                          2     2     4     4     8         8




Jeff Bivin -- LZHS
                     Find all real zeros
            f ( x)  8 x  2 x  21x  7 x  3
                         4         3             2

      x=1
                             8          2      -21           -7    3
                     1                  8        10         -11   -18
                             8         10      -11          -18   -15


                                 Remainder ≠ 0

                                 Therefore, not a factor.



Jeff Bivin -- LZHS
                     Find all real zeros
            f ( x)  8 x  2 x  21x  7 x  3
                         4         3             2

      x=3
                             8          2      -21           -7     3
                     3                 24        78         171   492
                             8         26        57         164   495


                                 Remainder ≠ 0

                                 Therefore, not a factor.



Jeff Bivin -- LZHS
                     Find all real zeros
            f ( x)  8 x  2 x  21x  7 x  3
                          4           3           2

 x  23
                                  8       2   -21      -7   3
x 3 0
    2
                     3               -12         15   9    -3
                      2
                                  8   -10         -6   2    0

                      x   8 x  10 x  6 x  2 
                              3
                              2
                                      3       2


                     x  2  2  4 x  5 x  3 x  1 
                           3         3     2



Jeff Bivin -- LZHS
                     2 x  3 4 x  5x  3x  1 
                                     3     2
                      Find all real zeros
            f ( x)  8 x  2 x  21x  7 x  3
                            4       3           2


                                    
             f ( x)  2 x  3 4 x  5 x  3x  1
                                           3         2
                                                               
                     1
                     3                     1, 3
                       =
                     8
                     4                  1, 2, 4, 8

                        1     3     1     3      1         3
       1  3                                       
                        2     2     4     4      8         8



Jeff Bivin -- LZHS
                      Find all real zeros
                          4 x  5 x  3x  1
                            3       2

 x 14
                                4       -5          -3   1
x 1 0
   4
                      1                     1       -1   -1
                      4
                                4       -4          -4   0

                     2 x  3 x    4 x  4 x  4 
                                        1
                                        4
                                                2


                     2 x  3 x  4  4  x  x  1 
                                     1        2


                     2 x  3 4 x  1  x  x  1 
                                            2

Jeff Bivin -- LZHS
                          Find all real zeros
                     f ( x)  8 x  2 x  21 x  7 x  3
                                   4           3             2


                     2 x  3 4 x  1  x       2
                                                        x 1  0   
              2x  3  0        4x  1  0             x2  x  1  0
                   x  23           x 14                 x
                                                                 1 1 4 (1)(1)
                                                                       2

                                                           x    1 5
                                                                   2


                                  3
                                    2
                                         1
                                        , ,
                                         4
                                              1 5
                                                2
                                                       
                           Rational Rational Irrational
Jeff Bivin -- LZHS                           but Real
                         Look at the graph
                     f ( x)  8 x  2 x  21 x  7 x  3
                               4             3          2


                                 3
                                   2
                                        1
                                       , ,
                                        4
                                             1 5
                                               2
                                                           ≈ 1.618
                                                            ≈ -0.618




Jeff Bivin -- LZHS
                         Look at the graph
                     f ( x)  8 x  2 x  21 x  7 x  3
                               4      3      2




Jeff Bivin -- LZHS
                       Find all real zeros
                     f ( x)  x  16 x  40 x  25
                                4         2



                     25
                     p                 1, 5, 25
                            factor of constant term a0
                        =
                     q1   factor of leading coefficient an
                                           1


                     1  5    25




Jeff Bivin -- LZHS
                      Find all real zeros
                f ( x)  x  16 x  40 x  25
                          4           2

      x=1
                              1       0   -16   -40   -25
                      1               1     1   -15   -55
                              1       1   -15   -55   -80
      x=5
                      5               5    25    45    25
      x–5=0
                              1       5     9     5     0


                                  
                 f ( x )   x  5 x  5 x  9 x  5
                                      3     2
                                                        
Jeff Bivin -- LZHS
                       Find all real zeros
                f ( x)  x  5 x  5 x  9 x  5
                                   3     2

      x = -1
                            1     5        9    5
    x+1=0            -1          -1       -4   -5
                            1     4        5    0

                                      
             f ( x)  x  5 x  1 x  4 x  5
                                           2
                                                       

Jeff Bivin -- LZHS
                      Find all real zeros
                 f ( x)  x  16 x  40 x  25
                            4          2


                                           
       f ( x)  x  5 x  1 x  4 x  5  02
                                                                    
               x50       x 1 0             x2  4x  5  0
                 x5          x  1                x
                                                         4  16  4 (1)(5 )
                                                                2
                                                         4  4
                                                    x
                      5,  1 
                                                             2
                                                         4  2i
                                                    x      2

                                                     Imaginary
                                                     -- not Real
Jeff Bivin -- LZHS
                      Look at the graph
                 f ( x)  x  16 x  40 x  25
                         4       2




        Note:
        x-min: -10
        x-max: 10
        x-scale: 1
        y-min: -250
        y-max: 100
        y-scale: 50



Jeff Bivin -- LZHS
                       Find all real zeros
    f ( x)  x  3x  5 x  15 x  4 x  12
                      5       4       3          2



                     12
                     p      factor1, 2, 3, 4, 6, 12 a0
                                   of constant term
                        =
                     q1   factor of leading coefficient an
                                          1


                     1  2   3 4        6  12




Jeff Bivin -- LZHS
                     Find all real zeros
    f ( x)  x  3x  5 x  15 x  4 x  12
                     5       4            3        2


      x=1
                             1       -3       -5   15       4   -12
    x-1=0                1            1       -2   -7       8    12
                             1       -2       -7    8      12     0


                             
 f ( x)  x  1 x  2 x  7 x  8 x  12
                                 4            3        2
                                                                      

Jeff Bivin -- LZHS
             Find all real zeros
 f ( x)  x  1  x  2 x  7 x  8 x  12 
                     4     3     2


      x=2
                         1   -2       -7     8    12
    x-2=0            2        2        0   -14   -12
                         1    0       -7    -6     0


                                  
         f ( x)  x  1 x  2 x  7 x  6
                                      3
                                                 

Jeff Bivin -- LZHS
                  Find all real zeros
        f ( x)  x  1 x  2  x  7 x  6 
                                    3


      x=3
                          1      0           -7    -6
    x-3=0            3           3            9     6
                          1      3            2     0


                                     
   f ( x)  x  1 x  2 x  3 x  3x  2
                                         2
                                                        

Jeff Bivin -- LZHS
                     Find all real zeros
    f ( x)  x  3x  5 x  15 x  4 x  12
                     5    4       3        2


                                          
   f ( x)  x  1 x  2 x  3 x  3x  2 2
                                                     
  f ( x)  x  1 x  2 x  3 x  2  x  1 
 x 1 0 x  2  0            x30 x20    x 1 0
    x 1      x2               x3   x  2    x  1

                          1, 2, 3,  2,  1 

Jeff Bivin -- LZHS
                      Look at the graph
    f ( x)  x  3x  5 x  15 x  4 x  12
                     5      4      3         2

                           1, 2, 3,  2,  1 
          End Behavior?




Jeff Bivin -- LZHS
                     Look at the graph
    f ( x)  x  3x  5 x  15 x  4 x  12
                     5     4      3         2

                          1, 2, 3,  2,  1 

        Note:
        x-min: -5
        x-max: 5
        x-scale: 1
        y-min: -20
        y-max: 20
        y-scale: 5



Jeff Bivin -- LZHS
        Fundamental Theorem of Algebra

                     If f(x) is a polynomial function of
                     degree n where n>0, then the
                     equation f(x) = 0 has at least one
                     solution in the set of complex
                     numbers.




Jeff Bivin -- LZHS
               Corollary to the Fundamental
                   Theorem of Algebra
                     If f(x) is a polynomial function of
                     degree n where n>0, then the
                     equation f(x) = 0 has exactly n
                     solutions provided each solution
                     repeated twice is counted as 2
                     solutions, each solution repeated
                     three times is counted as 3
                     solutions, and so on.

Jeff Bivin -- LZHS
                          Find all real zeros
    f ( x)  x  3x  6 x  10 x  21 x  9
                      5       4       3        2



                     p
                     9                1, 3, 9
                           factor of constant term a0
                       =
                     q
                     1   factor of leading coefficient an
                                         1

                              1  3  9


Jeff Bivin -- LZHS
                         Find all real zeros
    f ( x)  x  3x  6 x  10 x  21 x  9
                     5           4         3            2


   f ( x)  x  1 x  1 x  1 x  3  x  3 
                     x 1 0                   x30
                        x 1                     x  3
                                      1,  3 
                         1 is a multiple root with multiplicity 3
                     -3 is a multiple root with multiplicity 2


Jeff Bivin -- LZHS
                         Find all real zeros
 f ( x)  x  6 x  7 x  16 x  10 x  20
                     5       4        3        2



                     20
                     p           1, 2, 4, 5, 10, 20
                            factor of constant term a0
                        =
                     q1   factor of leading coefficient an
                                          1

                      1  2  4  5  10  20



Jeff Bivin -- LZHS
                         Find all real zeros
  f ( x)  x  6 x  7 x  16 x  18 x  20
                     5       4            3        2


      x=1                    1       -6        7   16   -18   -20
                         1            1       -5    2    18     0
         x=2
                             1       -5        2   18     0   -20
   x-2=0                 2            2       -8   -2    28    20
                             1       -4       -1   14    10     0

                             
 f ( x)  x  2 x  4 x  x  14 x  10
                                 4            3    2
                                                                    
Jeff Bivin -- LZHS
              Find all real zeros
  f ( x)  x  2  x  4 x  x  14 x  10 
                      4     3   2


      x=2
                          1   -4       -1       14    10
                     2         2       -4       -10    8
                          1   -2       -5         4   18
         x = -1
   x+1=0             -1       -1       5        -4    -10
                          1   -5       4        10      0

                              
  f ( x)  x  2 x  1 x  5 x  4 x  10
                                   3        2
                                                            
Jeff Bivin -- LZHS
             Find all real zeros
 f ( x)  x  2 x  1  x  5 x  4 x  10 
                             3     2


      x = -1
                          1   -5    4       10
                     -1       -1    6       -10
                          1   -6   10         0
      x+1=0




                                   
   f ( x)  x  2 x  1 x  1 x  6 x  10
                                        2
                                                   
Jeff Bivin -- LZHS
              Find all real zeros
   f ( x)  x  2 x  1 x  1 x  6 x  10 
                                      2


       x  2 x 1 x 1 x        2
                                            6 x  10  0       
               x20       x 1 0         x 2  6 x  10  0
                 x2          x  1            x
                                                     6     36  4 (1)(10 )
                                                               2
                                                       6     4
                                                  x
                      2,  1 
                                                           2
                                                       6  2i
                                                  x      2

                                                  x3i

Jeff Bivin -- LZHS
  f ( x)  x  6 x  7 x  16 x  18 x  20
                        5       4        3        2

            x                               
                      2   x  1  x  1 x 2  6 x  10    0
                                      1, 2 
          End Behavior?




Jeff Bivin -- LZHS
                             Key Concepts

                     If f(x) is a polynomial function with
                     real coefficients, and a + bi is an
                     imaginary zero of f(x), then a - bi is
                     also a zero of f(x).
                       Imaginary solutions appear in
                             conjugate pairs.



Jeff Bivin -- LZHS
                         Key Concepts
              If f(x) is a polynomial function with
              rational coefficients, and a and b are
              rational numbers such that ---- is b
                               a b
              irrational. If --------- is a zero of f(x),
                      a b
              then --------- is also a zero of f(x).
            Irrational solutions containing a square
                 root appear in conjugate pairs.


Jeff Bivin -- LZHS
    Write a polynomial function f(x) of least degree that
    has rational coefficients, leading coefficient of 1,
    and the following zeros  1, -2, 4.
                       x=1          x = -2        x=4
                     x–1=0       x+2=0         x–4=0

                      f(x) = (x – 1) (x + 2) (x – 4)
                      f(x) = (x – 1) (x2 – 4x + 2x – 8)
                      f(x) = (x – 1) (x2 – 2x – 8)
                      f(x) = x3 – 2x2 – 8x – x2 + 2x + 8
                      f(x) = x3 – 3x2 – 6x + 8

Jeff Bivin -- LZHS
    Write a polynomial function f(x) of least degree that
    has rational coefficients, leading coefficient of 1,
    and the following zeros  4,  1, 2  3
            x4        x  1     x2 3
    x40            x 1 0    x2 30

          f ( x)    x  4 x  1 x  2    3   
                          If 2  3 is a zero ,
                          then 2  3 is a zero!


Jeff Bivin -- LZHS
    Write a polynomial function f(x) of least degree that
    has rational coefficients, leading coefficient of 1,
    and the following zeros  4,  1, 2  3
            x4        x  1     x2 3         x2 3
    x40            x 1 0    x2 30        x2 30


                                          
        f ( x)  x  4 x  1 x  2  3 x  2  3   

Jeff Bivin -- LZHS
                                        
      f ( x)  x  4 x  1 x  2  3 x  2  3  
    f ( x)  x  3x  4 x  2  3  x  2  3 
                      2



            f ( x)  x  3x  4 x  2  3
                                      2                  2



                             2
                                       
            f ( x)  x  3x  4 x  4 x  4  3
                                              2
                                                                 
                f ( x)       x  3x  4x
                                  2               2
                                                             
                                                       4x  1

Jeff Bivin -- LZHS
                             
                f ( x)  x  3x  4 x  4 x  1
                                 2
                                                        2
                                                              
                     x  4x  x
                     4               3       2


                          3x  12 x  3x
                                     3           2


                                4 x  16 x  4
                                    2


             f ( x)  x  7 x  9 x  13 x  4
                         4               3           2




Jeff Bivin -- LZHS
    Write a polynomial function f(x) of least degree that
    has rational coefficients, leading coefficient of 1,
    and the following zeros  2, 3  2i
                      x2                  x  3  2i
               x20                   x  3  2i  0

                     f ( x)    x  2 x  3  2i 
                                  If 3  2i is a zero,
                                  then 3  2i is a zero!


Jeff Bivin -- LZHS
    Write a polynomial function f(x) of least degree that
    has rational coefficients, leading coefficient of 1,
    and the following zeros  2, 3  2i
                      x2                  x  3  2i         x  3  2i
               x20                   x  3  2i  0     x  3  2i  0

                     f ( x)    x  2 x  3  2i  x  3  2i 




Jeff Bivin -- LZHS
          f ( x)           x  2 x  3  2i  x  3  2i 
    f ( x)               x  2 [ x  3]  2i  [ x  3]  2i 
                                         
                        f ( x)  x  2 x  3  4i
                                                    2
                                                        
                                                        2


                         f ( x)  x  2 x  3  4
                                                    2


                                        
                     f ( x)   x  2 x  6 x  9  4
                                            2
                                                           
                      f ( x)     x  2   x  6 x  13 
                                                2


Jeff Bivin -- LZHS
                                         
                     f ( x)   x  2  x  6 x  13
                                             2
                                                       
                            x  6 x  13 x
                             3       2


                                  2 x  12 x  26
                                     2



                     f ( x)  x  8 x  25 x  26
                             3       2




Jeff Bivin -- LZHS
                     Descartes’ Rule of Signs
      Let f(x) = anxn + an-1xn-1 +. . . + a1x + a0 be a
      polynomial function with real coefficients.
       The number of positive real zeros of f(x) is
      equal to the number of changes in sign of the
      coefficients of f(x) or is less than this by an even
      number.
       The number of negative real zeros of f(x) is
      equal to the number of changes in sign of the
      coefficients of f(-x) or is less than this by an even
      number.
Jeff Bivin -- LZHS
     Determine the number of positive real zeros.
  f ( x)  x  6 x  7 x  16 x  18 x  20
                     5   4        3         2




                     1    2                     3
         3 sign changes 
                    f(x) has 3 or 1 positive real zero(s).




Jeff Bivin -- LZHS
    Determine the number of negative real zeros.
  f ( x)  x  6 x  7 x  16 x  18 x  20
                     5       4       3           2

   f ( x)   x   6 x   7 x   16  x   18  x   20
                   5         4         3           2


      f ( x)   x  6 x  7 x  16 x  18 x  20
                         5       4       3       2




                                             1           2
         2 sign changes 
                    f(x) has 2 or 0 negative real zero(s).


Jeff Bivin -- LZHS
                             Putting it together !
  f ( x)  x  6 x  7 x  16 x  18 x  20
                         5       4          3               2

          f(x) has 3 or 1 positive real zero(s)
          f(x) has 2 or 0 negative real zero(s)
                 Positive       Negative        Imaginary       Total
                real zeros     real zeros         zeros         zeros
                     3               2             0             5
                     3               0             2             5
                     1               2             2             5
                     1               0             4             5
Jeff Bivin -- LZHS
                         Look at the graph
  f ( x)  x  6 x  7 x  16 x  18 x  20
                     5      4    3     2




        Note:
        x-min: -5
        x-max: 5
        x-scale: 1
        y-min: -30
        y-max: 20
        y-scale: 5



Jeff Bivin -- LZHS
     Determine the number of positive real zeros.
 f ( x)  12 x  6 x  7 x  6 x  2 x  20
                     7      5        4       2




         0 sign changes 
                    f(x) has 0 positive real zero(s).




Jeff Bivin -- LZHS
    Determine the number of negative real zeros.
 f ( x)  12 x  6 x  7 x  6 x  2 x  20
                         7       5       4       2

  f ( x)  12  x   6 x   7 x   6 x   2 x   20
                     7         5         4         2


f (  x)                                               


                                 1               2       3
         3 sign changes 
                    f(x) has 3 or 1 negative real zero(s).



Jeff Bivin -- LZHS
                             Putting it together !
 f ( x)  12 x  6 x  7 x  6 x  2 x  20
                             7           5     4          2

          f(x) has 0 positive real zero(s)
          f(x) has 3 or 1 negative real zero(s)
                 Positive         Negative    Imaginary       Total
                real zeros       real zeros     zeros         zeros
                     0               3             4           7
                     0               1             6           7




Jeff Bivin -- LZHS
                      Look at the graph
 f ( x)  12 x  6 x  7 x  6 x  2 x  20
                      7    5    4    2




        Note:
        x-min: -5
        x-max: 5
        x-scale: 1
        y-min: -50
        y-max: 50
        y-scale: 10



Jeff Bivin -- LZHS
                       Find all real zeros
                 f ( x)  18 x  63 x  40 x  21
                                   3             2


                     21
                     p      factor of1, 3, 7, 21 term a0
                                      constant
                        =
                     q
                     18   factor of leading coefficient an
                                  1, 2, 3, 6, 9, 18

             1  3  7  21                   1      3     7   21
                                                   2      2     2   2
            3 3 6 6    
              1   7    1   7                      1
                                                  9
                                                          7
                                                          9
                                                                1
                                                               18
                                                                     7
                                                                    18
                      f(x) has 2 or 0 positive real zero(s)
                      f(x) has 1 negative real zero(s)


Jeff Bivin -- LZHS
                             Find all real zeros
                 f ( x)  18 x  63 x  40 x  21
                                             3                2



       21  21  7  7
             2        2                                       3  7  3  7
                                                                   3   2   6

          1                7
                             9          1
                                         2
                                                     7
                                                     18    1  1  1  18
                                                            3   6   9
                                                                         1

           1             1           1               1         7   1   7
          18             9           6               3        18   2   9    1
        7            3           7                        7            21
        6            2           3           3            2        7   2    21

Jeff Bivin -- LZHS
                       Find all real zeros
                     f ( x)  18 x  63 x  40 x  21
                                  3         2

                             18       -63       40     21
                       1               18       -45     -5
                             18       -45        -5     16
                       2               36       -54    -28
                             18       -27       -14     -7
                       3               54       -27     39
                             18        -9        13     60
                       7              126       441   3367
                             18        63       481   3388

Jeff Bivin -- LZHS
                             Find all real zeros
                 f ( x)  18 x  63 x  40 x  21
                                             3                2



       21  21  7  7
             2        2                                       3  7  3  7
                                                                   3   2   6

          1                7
                             9          1
                                         2
                                                     7
                                                     18    1  1  1  18
                                                            3   6   9
                                                                         1

           1             1           1               1         7   1   7
          18             9           6               3        18   2   9    1
        7            3           7                        7            21
        6            2           3           3            2        7   2    21

Jeff Bivin -- LZHS
                                                      Location
                       Find all real zeros            Principle


                     f ( x)  18 x  63 x  40 x  21
                                  3         2

                             18       -63       40     21
                       1               18       -45     -5
                             18       -45        -5     16
                       2               36       -54    -28
                             18       -27       -14     -7
                       3               54       -27     39
                             18        -9        13     60
                       7              126       441   3367
                             18        63       481   3388

Jeff Bivin -- LZHS
                             Find all real zeros
                 f ( x)  18 x  63 x  40 x  21
                                             3                2



       21  21  7  7
             2        2                                       3  7  3  7
                                                                   3   2   6

          1                7
                             9          1
                                         2
                                                     7
                                                     18    1  1  1  18
                                                            3   6   9
                                                                         1

           1             1           1               1         7   1   7
          18             9           6               3        18   2   9    1
        7            3           7                        7            21
        6            2           3           3            2        7   2    21

Jeff Bivin -- LZHS
                           Find all real zeros
                     f ( x)  18 x  63 x  40 x  21
                                    3         2


                               18       -63       40    21
                       3
                       2                 27       -54   -21
                               18       -36       -14     0
                       7
                       3                 42        14
                               18         6         0

                               18x  6  0
                                  18x  6
                                     x 13

Jeff Bivin -- LZHS

				
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Description: how to make a solution of the formula of Polynomial