The document explains the Rational Root Theorem and its application in finding possible rational roots of polynomial equations with integer coefficients. It provides examples of determining rational zeros through synthetic division and outlines concepts like the Fundamental Theorem of Algebra and Descartes' Rule of Signs for analyzing the number of positive and negative real roots. Various polynomial functions are explored to illustrate the methods of finding their roots.

1. Rational Root Theorem

2. Rational Root Theorem:
Suppose that a polynomial equation with
integral coefficients has the root , where
h and k are relatively prime integers. Then
h must be a factor of the constant term of
the polynomial and k must be a factor of
the coefficient of the highest degree term.
(useful when solving higher degree polynomial
equations)
h
k

3. How many solutions does the equation have?
1. 4 2
5 36 0
x x
  
4
2. 3 2
5 7 8 16 0
x x x
   
3

4. a) State all possible rational zeros for
g(x) = x3
+ 2x2
- 3x + 5
Possible rational zeros: +5, +1
b) State all possible rational zeros
for g(x) = 6x3
+ 6x2
- 15x - 2
+6,+3,+2,+1 +2,+1
+1 +5,+1
Factors of the constant
Factors of the leading coefficient
Possible rational zeros:
+2,+1
+6,+3,+2,+1 2
1
6
1
3
2
3
1
,
,
2
,
1
,
, 





Example 1

5. Solve using the Rational Root
Theorem:
 4x2
+ 3x – 1 = 0 (any rational root must have a numerator
that is a factor of -1 and a denominator
that is a factor of 4)
factors of -1: ±1
factors of 4: ±1,2,4
possible rational roots: (now use synthetic
division
to find rational roots)
1 1
1, ,
2 4

1 4 3 -1
4 7
4 7 6 no
-1 4 3 -1
-4 1
4 -1 0 !
yes
4 1 0
4 1
1
4
x
x
x
 


1
1,
4
x 
(note: not all possible rational roots are roots!)

6. Ex : Solve using the Rational Root
Theorem:3 2
2 13 10 0
x x x
   
: 1,2,5,10
: 1
h
k


1 1 2 -13 10
1 3 -10
1 3 -10 0 !
yes
  
2
3 10 0
5 2 0
5, 2
x x
x x
x
  
  

5,1, 2
x 
1, 2, 5,10

possible rational roots:

7. Ex : Solve using the Rational Root
Theorem: 3 2
4 4 0
x x x
   
: 1,2,4
: 1
h
k


possible rational roots: 1, 2, 4

1 1 -4 -1 4
1 -3 -4
1 -3 -4 0 !
yes
  
2
3 4 0
4 1 0
1, 4
x x
x x
x
  
  

1,1, 4
x 

8. Ex : Solve using the Rational Root
Theorem: 3 2
3 5 4 4 0
x x x
   
: 1,2,4
: 1, 3
h
k


possible rational roots:
1 2 4
1, 2, 4, , ,
3 3 3

-1 3 -5 -4 4
-3 8
3 -8 -4
-4
0 !
yes
  
2
3 8 4 0
3 2 2 0
2
, 2
3
x x
x x
x
  
  

2
1, , 2
3
x 
To find other roots can use synthetic division
using other possible roots on these coefficients.
(or factor and solve the quadratic equation)
2 3 -8 4 3 2 0
6 -4 3 2
3 -2 0
x
x
 

2
3
x 

9. Find all zeros of the polynomial function.
3.
4 3 2
( ) 4 3 4 4
f x x x x x
    
Factors of last term:
Factors of first term: 1
Possible rational zeros:
How many answers: 4
1, 2, 4
1, 2, 4

10. 1 1 –4 3 4
1
1
–3
–3
0
0
4
Find all real zeros of the function.
3.
Possible rational zeros: 1, 2, 4
4 3 2
( ) 4 3 4 4
f x x x x x
    
–4
4
0
(x – 1) x3
– 3x2
+ 0x + 4
( ) = 0

11. 1 1 –4 3 4
1
1
–3
–3
0
0
4
Find all real zeros of the function.
3.
Possible rational zeros: 1, 2, 4
4 3 2
( ) 4 3 4 4
f x x x x x
    
–4
4
0
–1
1
–1
–4
4
4
–4
0
(x + 1) x2
– 4x + 4
( ) = 0
(x – 1)
x
x
–2
–2
(x – 1)(x + 1)(x – 2)(x – 2) = 0
x = 1, 2

12. Find all zeros of the polynomial function.
4.
4 3 2
( ) 7 9 18
f x x x x x
    
1
How many answers: 4
1, 18
2, 9,
3, 6,
1, 18
2, 9,
3, 6,
Factors of last term:
Factors of first term:
Possible rational zeros:

13. 1 1 –1 7 –9
1
1
0
0
7
7
–2
Find all real zeros of the function.
4.
Possible rational zeros:

4 3 2
( ) 7 9 18
f x x x x x
    
1, 18
2, 9,
3, 6,
–18
–2
–20

14. –1 1 –1 7 –9
1
–1
–2
2
9
–9
–18
Find all real zeros of the function.
4.
Possible rational zeros:
4 3 2
( ) 7 9 18
f x x x x x
    
1, 18
2, 9,
3, 6,
–18
18
0
(x + 1)
(x + 1)(x2
+ 9)(x – 2) = 0
x3
– 2x2
+ 9x – 18
( ) = 0
( ) ( )
x2
(x – 2) +9 (x – 2)
(x2
+ 9) (x – 2)
(x + 1)
x = 1, 3 , 2
i
 

15. Find the roots of x3
+ 8x2
+ 16x + 5 = 0

16. Ex : Determine the zeros:
3 2
3 2 6
x x x
  

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

18. Complex Conjugates Theorem
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 numbers always come in pairs)

19. Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1, and given zeros.
5. 3, –2
f(x) = (x – 3)(x + 2)
x2
+ 2x – 3x – 6
f(x) = x2
– x – 6

20. Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1, and given zeros.
6. i, –2i
f(x) = (x – i)(x + i)(x + 2i)(x – 2i)
f(x) = x4
+ 5x2
+ 4
(x2
– i2
) (x2
– 4i2
)
(x2
+ 1) (x2
+ 4)
x4
+ 4x2
+ x2
+ 4

21. Some possibilities for zeros (roots)
Equation Real Zeros Imaginary Zeros
Linear 1 0
Quadratic 2 0
0 2
Cubic 3 0
1 2
Quartic 4 0
2 2
0 4
Quintic 5 0
3 2
1 4

22. Find the zeros.
f(x) = x 3
– x2
+ 4x – 4 = 0
g(x) = x 4
+ 2x 3
– 5x 2
– 4x + 6 = 0

23. Write a polynomial function of least degree with a leading coefficient of
1.
2, 3, -4
(x – 2)(x – 3)(x + 4) = 0
(x2
– 5x + 6)(x + 4) = 0
x 3
- x2
– 14x + 24 = 0

24. Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1, and given zeros.
7.
4, 2
   
( ) 4 2 2
f x x x x
   
  
2
4 2
x x
 
x3
– 2x – 4x2
+ 8
f(x) = x3
– 4x2
– 2x + 8

25. Write a polynomial function of least degree with a leading coefficient of
1.
7, i
(x – 7)(x – i)(x + i) = 0
(x – 7)(x 2
+ 1) = 0
x 3
– 7 x 2
+ x – 7 = 0
ASSIGNMENT:

26. QUIZ: Find the roots of the following:
1. x4
+ 5x3
- 4x2
- 4x - 5 = 0
2. x3
+ x2
+ 2x - 1 = 0
3. x2
+ 5x - 6 = 0
1. 1, 2, 3
2. -3, i
Write a polynomial function of least degree with a leading coefficient of
1.

27. Descartes’ Rule of Signs
Let f(x) be a polynomial function with
real coefficients.
 The number of positive real zeros of f(x) is
equal to the number of sign changes 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 sign changes
of the coefficients of f(-x) or is less than
this by an even number.

28. huh
? ? ?

29. Using Descartes’ Rule of Signs….
f(x) = x6
-2x 5
+3x 4
– 10x 3
-6x 2
-8x -8
3 sign changes, so f has 3 or 1 positive real roots
f(-x) = x6
+2x 5
+3x 4
+ 10x 3
-6x 2
+8x -8
3 sign changes, so f has 3 or 1 negative real roots

30. g(x) = -x5
– 5x4
+ 7x3
– 4x2
– 8x + 9
 Determine the possible number of
positive real zeros, negative real zeros,
and imaginary zeros of the function
given above.
 g(x) = -x5
– 5x4
+ 7x3
– 4x2
– 8x + 9
 g(-x) = x5
– 5x4
- 7x3
– 4x2
+ 8x + 9

31. Example
 Determine the possible number of positive
real zeros, negative real zeros, and
imaginary zeros of the function given
above.
3
2
3
4
3 2
3
4
6





 x
x
x
x
x
x
f )
(
3
2
3
4
3 2
3
4
6











 )
(
)
(
)
(
)
(
)
(
)
( x
x
x
x
x
x
f
3
2
3
4
3 2
3
4
6






 x
x
x
x
x
x
f )
(

32. Use Descartes’ Rule of signs to find the number of possible real
roots.
3 2
( ) 5 4 12
f x x x x
   