# 2.2b Rational zeros of Polynomial Functions by xiaoyounan

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```									              Rational zeros of a polynomial
Consider the polynomial function: P(x) = 8x2 – 2x – 3
What are the zeros of P(x)?                        4x              2x
8x2 – 2x – 3 = (4x – 3)(2x + 1)
3
4x – 3 = 0 gives:        x
4                             -3      +1
1
2x + 1 = 0 gives:       x
3            1           2
x  and x          are the rational zeros of P(x)
4            2
The numerators 3 and -1 of these zeros are factors of the last term -3 of P(x)
The denominators 4 and 2 of these zeros are factors of the leading
coefficient 8 of P(x)
We can use this to make a general statement about rational zeros.

If b/c is a rational zero of a polynomial, then b must be
an integer factor of the constant term and c must be an
integer factor of the leading coefficient.
Find all rational zeros of: P(x) = 2x3 – 9x2 + 14x – 5
The factors of -5 are -5 and 1 or 5 and -1
 5 ,  1 are the possible value of b
The factors of 2 are 2 and 1 or -2 and -1
 2,  1 are possible values of c
The possible rational zeros are of the form b/c
What are the possible values of b/c?
5       1
b/c can have values: x  1, 5,    and 
2       2
We will use TI84 to figure out which of these
represent the rational zeros of P(x)
1
x  is a rational zero of P(x)
2
To obtain the complex zeros, we do synthetic division of P(x)
using the zero x = ½ and obtain the remaining factor.
Find all rational zeros of: P(x) = 2x3 – 9x2 + 14x – 5
2        -9       14           -5
The remaining factor is:                     1       -4        5
1
2x2 – 8x + 10                      2       -8      10         0
2
To find the complex zeros, solve 2x2 – 8x + 10 = 0
a = 2, b = -8 and c = 10

b  b 2  4ac (8)  64  4  2  10   8  64  80
x                                       
2a                 22                 4
8  16      8  4i 4(2  i)  2  i
                   
4          4       4
x = 2 + i is a zero.    [x – (2 + i )] is a factor of P(x)

x = 2 – i is a zero.    [x – (2 – i )] is a factor of P(x)

Challenge: Show that [x – (2 + i)][x – (2 – i)] = 2x2 – 8x + 10
Write P(x) = 3x3 – 10x2 + 31x +26 as the product of two factors, given that
x = -2/3 is a zero.                  3      -10      31       26
-2        8     -26
2                             2
x  is a linear factor          3         -12       39       0
3                               3
Use synthetic division to
find the other linear factor.
The other factor is: 3x2 – 12x + 39

    2
P( x)   x   (3x 2  12 x  39)
    3
 3x  2  2
           (3x  12 x  39)
 3  2
 3 x  12 x  39                3x 2 12 x 39 
  3x  2                      3x  2            
        3                       3     3    3 
  3x  2  x2  4 x  13
Find all roots of the equation:
x4 – 2x2 – 16x – 15 = 0.
The roots of the equation are the values
of x that make the equation true.
They are the zeros of the corresponding polynomial function.
We start by finding the real zeros on TI84.
x = -1 is a real zero
Use synthetic division to find the remaining factor.
The remaining factor is:
1      0     -2     -16 -15
x 3 – x2 – x – 15
-1      1      1   15
x = 3 is a real zero          -1 1     -1     -1     -15   0
Use synthetic division to find the remaining factor.
Remaining factor is:
1    -1       -1    -15
x2 +2x + 5                             3       6     15
3 1       2       5      0
Find all roots of the equation: x4 – 2x2 – 16x – 15 = 0.
Real roots: x = -1, x = 3         Remaining factor: x2 +2x + 5
Solve x2 + 2x + 5 = 0 using quadratic formula.
a = 1, b = 2 and c = 5

b  b 2  4ac    2  4  4 1  5     2  4  20
x                                     
2a                 2 1                 2
2  16        2  4i 2(1  2i )
                        
2              2         2
 1  2i
The roots of the equation are:

x = -1, 3, -1 + 2i and -1 – 2i
13 2 5         1
Find all zeros of:   P( x )  x 4    x  x
4      2      4
Multiplying P(x) by a constant will not change its zeros. Can you say why?
Multiplying by 4 gives:
P(x) = 4x4 – 13x2 – 10x – 1
We start by finding the real zeros on TI84.
x = -1 is an integer zero of multiplicity 2
The other two zeros are irrational
Do synthetic division twice
using the zero x = -1 to find
4       0      -13     -10     -1
the remaining factor
-4        4       9     1
The remaining factor is:
4x3 – 4x2 – 9x – 1             -1 4       -4       -9      -1     0
repeat synthetic division           4        -4      -9      -1
using x = -1
-4       8       1
Remaining factor is:
-1    4        -8      -1       0
4x2 – 8x – 1
Real roots: x = -1, x = -1 Remaining factor: 4x2 – 8x – 1
To find the irrational zeros we solve: 4x2 – 8x – 1 = 0
a = 4, b = -8 and c = -1

b  b 2  4ac   (8)  64  4  4  (1)
x                
2a                  24
8  64  16    8  80 8  16  5
                     
8           8       8



8 4 5 4 2 5


8          8



2 5  
2      2 5
One irrational zero is:   x              -0.118
2
2 5
Second irrational zero is: x          2.118
2
Write the polynomial P(x) = x3 – 8x2 + 17x – 4 as the product of linear factors.
Start by finding real integer zeros
x = 4 is a real integer zero
x – 4 is a linear factor
Find the remaining factor by synthetic division:
The remaining factor is:
x2 – 4x + 1                           1      -8            17      -4
This has two irrational zeros                 4            -16      4
and we use the quadratic           4 1       -4             1       0
formula to find them.
a = 1, b = -4 and c = 1

b  b 2  4ac (4)  16  4  1  1 4  16  4   4  12
x                                                
2a              2 1                2           2

4  43 4  2 3 2 2  3
       
       
 2 3
2         2          2
Write the polynomial P(x) = x3 – 8x2 + 17x – 4 as the product of linear factors.
x – 4 is a linear factor

x  2  3 are irrational zeros

x  2  3 is a irrational zero
The corresponding factor is:                   
x  2  3   x  2  3
                                
x  2  3 is an irrational zero
The corresponding factor is:                   
x  2  3   x  2  3



x 3  8 x 2  17 x  4  ( x  4) x  2  3 x  2  3                 
The polynomial P(x) = x4 – 4x3 + 3x2 + 8x – 10 has a
zero x = 2 + i. Find all other zeros
x = 2 + i is a zero means that x = 2 – i is also a zero. Why?
Do synthetic            1        –4         3                 8           –10
division on the
2+i        -5              -4 – 2i        10
result using x =
2+i 1        -2 + i      -2               4 – 2i         0
2 + i and also x
=2–i
1       –2+i           2            4 – 2i
2–i           0          -4 + 2i

2–i 1           0           -2              0

The remaining factor is: x2 – 2                2, 0            2, 0   
x2 – 2 = 0 gives:  x 2
Find all zeros of P(x) = 6x4 + 35x3 + 2x2 – 233x – 360
We check for real integer zeros on TI84
There are no real integer zeros
There are two rational zeros. Can
you figure out what they are?
There are two complex zeros.
Using the values from TI83 we can conclude
that x = -9/2 and x = 8/3 are the rational zeros.
We do synthetic division
using the first zero x = -9/2      6     35     2        -233    -360
The remaining factor is:       9       -27    -36        153     360
6      8    -34         -80     0
6x3 + 8x2 – 34x – 80            2
repeat synthetic                    6      8      -34     -80
division using x = 8/3
16      64      80
Remaining factor is:          8
6      24      30       0
6x 2 + 24x + 30
3
Find all zeros of P(x) = 6x4 + 35x3 + 2x2 – 233x – 360
Solve 6x2 + 24 x + 30 = 0 for complex zeros.
6(x2 + 4x + 5) = 0
x2 + 4x + 5 = 0
a = 1, b = 4 and c = 5

b  b 2  4ac   4  16  4 1 5      4  16  20
x                                      
2a                  2 1                2
4  4        4  2i
              
2           2
2  2  i 
               2  i
2
9 8
The four zeros are:       x   , , 2  i and 2  i
2 3

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