# Polynomial by ainnaku

VIEWS: 8 PAGES: 69

• pg 1
```									          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

```
To top