1. 6.5 Theorems About Roots of Polynomial Equations
Rational Roots

2. POLYNOMIALS and THEOREMS Theorems of Polynomial Equations
There are 4 BIG Theorems to know about Polynomials
Rational Root Theorem
Irrational Root Theorem
Imaginary Root Theorem
Descartes Rule

3. Consider the following . . .
x3 – 5x2 – 2x + 24 = 0
This equation factors to:
(x+2)(x-3)(x-4)= 0
The roots therefore are: -2, 3, 4

4. Take a closer look at the original equation and our roots:
x3 – 5x2 – 2x + 24 = 0
The roots therefore are: -2, 3, 4
What do you notice?
-2, 3, and 4 all go into the last term, 24!

5. Spooky! Let’s look at another
24x3 – 22x2 – 5x + 6 = 0
This equation factors to:
(x+1)(x-2)(x-3)= 0
(Some factors of 24)
The roots therefore are: -1/2, 2/3, 3/4

6. Take a closer look at the original equation and our roots:
24x3 – 22x2 – 5x + 6 = 0
This equation factors to:
(x+1)(x-2)(x-3)= 0
(Some factors of 24)
The roots therefore are: -1, 2, 3
What do you notice?
The numerators 1, 2, and 3 all go into the last term, 6!
The denominators (2, 3, and 4) all go into the first term, 24!

7. This leads us to the Rational Root Theorem
For a polynomial,
If p/q is a root of the polynomial,
then p is a factor of an
and q is a factor of ao

8. Example (RRT) ±3, ±1 ±1 ±12, ±6 , ±3 , ± 2 , ±1 ±4 ±1 , ±3
1. For polynomial
Here p = -3 and q = 1
Factors of -3
Factors of 1
±3, ±1 ±1 
Or 3,-3, 1, -1
Possible roots are ___________________________________
2. For polynomial
Here p = 12 and q = 3
Factors of 12
Factors of 3
±12, ±6 , ±3 , ± 2 , ±1 ±4
±1 , ±3 
Possible roots are ______________________________________________
Or ±12, ±4, ±6, ±2, ±3, ±1, ± 2/3, ±1/3, ±4/3
Wait a second Where did all of these come from???

9. Let’s look at our solutions
±12, ±6 , ±3 , ± 2 , ±1, ±4
±1 , ±3
Note that + 2 is listed twice; we only consider it as one answer
Note that + 1 is listed twice; we only consider it as one answer
Note that + 4 is listed twice; we only consider it as one answer
That is where our 9 possible answers come from!

10. Let’s Try One
Find the POSSIBLE roots of 5x3-24x2+41x-20=0

11. Let’s Try One
5x3-24x2+41x-20=0

12. That’s a lot of answers!
Obviously 5x3-24x2+41x-20=0 does not have all of those roots as answers.
Remember: these are only POSSIBLE roots. We take these roots and figure out what answers actually WORK.

13. Step 1 – find p and q
p = -3
q = 1
Step 2 – by RRT, the only rational root is of the form…
Factors of p
Factors of q

14. Step 3 – factors
Factors of -3 = ±3, ±1
Factors of 1 = ± 1
Step 4 – possible roots
-3, 3, 1, and -1

15. Step 6 – synthetic division
Step 5 – Test each root
Step 6 – synthetic division
X X³ + X² – 3x – 3 -1 -3 3 1 -1
(-3)³ + (-3)² – 3(-3) – 3 = -12
(3)³ + (3)² – 3(3) – 3 = 24 3 -1
(1)³ + (1)² – 3(1) – 3 = -4 1 -3
(-1)³ + (-1)² – 3(-1) – 3 = 0
THIS IS YOUR ROOT
BECAUSE WE ARE LOOKING
FOR WHAT ROOTS WILL
MAKE THE EQUATION =0
1x² + 0x -3

16. Step 7 – Rewrite
x³ + x² - 3x - 3
= (x + 1)(x² – 3)
Step 8– factor more and solve
(x + 1)(x² – 3)
(x + 1)(x – √3)(x + √3)
Roots are -1, ± √3

17. Let’s Try One Find the roots of 2x3 – x2 + 2x - 1
Take this in parts. First find the possible roots. Then determine which root actually works.

18. Let’s Try One
2x3 – x2 + 2x - 1

20. Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
Step 1 – find p and q
p = -6
q = 1
Step 2 – by RRT, the only rational root is of the form…
Factors of p
Factors of q

21. Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
Step 3 – factors
Factors of -6 = ±1, ±2, ±3, ±6
Factors of 1 = ±1
Step 4 – possible roots
-6, 6, -3, 3, -2, 2, 1, and -1

22. Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
Step 6 – synthetic division
Step 5 – Test each root
X x³ – 5x² + 8x – 6 -6 6 3 -3 2 -2 1 -1
-450 78
-102 -2
-50
-20 3
THIS IS YOUR ROOT 6 3 -6 1 -2 2
1x² + -2x + 2

23. Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
Step 7 – Rewrite
x³ – 5x² + 8x – 6
= (x - 3)(x² – 2x + 2)
Step 8– factor more and solve
(x - 3)(x² – 2x + 2)
Roots are 3, 1 ± i
X= 3
Quadratic Formula

24. Complex ConjugateTheorem
For a polynomial
If a + bi is a root,
Then a – bi is also a root
Complex roots always come in pairs. Real values do not.
CONJUGATE ___________________________
Complex pairs of form a + bi and a - bi

25. Example 1. For polynomial has roots 3 + 2i 2 3 – 2i Other roots ______
Degree of Polynomial ______
2. For polynomial has roots -1, 0, - 3i, 1 + i 6
3i, 1 - i
Other roots __________
Degree of Polynomial ______

26. Irrational Root Theorem
For a polynomial
If a + √b is a root,
Then a - √b is also a root
Irrationals always come in pairs. Real values do not.
CONJUGATE ___________________________
Complex pairs of form a + √ b and a - √ b

27. Example (IRT) 1. For polynomial has roots 3 + √2 2 3 - √2
Other roots ______
Degree of Polynomial ______
2. For polynomial has roots -1, 0, - √3, 1 + √5 6
√3 , 1 - √5
Other roots __________
Degree of Polynomial ______

28. Example (IRT) 1. For polynomial has roots 1 + √3 and -√11 1 - √3 √11
Other roots ______ _______ 4
Degree of Polynomial ______
Question: One of the roots of a polynomial is
Can you be certain that is also a root?
No. The Irrational Root Theorem does not apply unless you know that all the coefficients of a polynomial are rational. You would have to have
as your root to make use of the IRT.

29. Write a polynomial given the roots 5 and √2
Another root is - √2
Put in factored form
y = (x – 5)(x + √2 )(x – √2 )

30. Decide what to FOIL first y = (x – 5)(x + √2 )(x – √2 )-2

31. FOIL or BOX to finish it up (x-5)(x² – 2)
y = x³ – 2x – 5x² + 10
Standard Form
y = x³ – 5x² – 2x + 10 x x -5 X3
-2x
-5x2 10

32. Write a polynomial given the roots -√5, √7
Other roots are √5 and -√7
Put in factored form
y = (x – √5 )(x + √5)(x – √7)(x + √7)
Decide what to FOIL first

33. y = (x – √5 )(x + √5)(x – √7)(x + √7)
Foil or use a box method to multiply the binomials
X √5
X √7 x √5 x √7 X2
-X √5 X2
-X √7
X √5 -5
X √7 -7
(x² – 5)
(x² – 7)

34. FOIL or BOX to finish it up y = x4 – 7x² – 5x² + 35 Clean up -7 X4
-5x2
-7x2 35