Solving All Polynomial Equations

1. Section 11-8
Solving All Polynomial Equations
Wednesday, March 18, 2009

2. Warm-up:
Solve for all possible solutions (real and imaginary).
1. 2x + 5 = 18 2. 4 -1=0
w
Wednesday, March 18, 2009

3. Warm-up:
Solve for all possible solutions (real and imaginary).
1. 2x + 5 = 18 2. 4 -1=0
w
x = 13/2
Wednesday, March 18, 2009

4. Warm-up:
Solve for all possible solutions (real and imaginary).
1. 2x + 5 = 18 2. 4 -1=0
w
(w2 - 1)(w2 +1) = 0
x = 13/2
Wednesday, March 18, 2009

5. Warm-up:
Solve for all possible solutions (real and imaginary).
1. 2x + 5 = 18 2. 4 -1=0
w
(w 2 - 1)(w2 +1) = 0
x = 13/2 (w2 - 1)(w2 - [-1]) = 0
Wednesday, March 18, 2009

6. Warm-up:
Solve for all possible solutions (real and imaginary).
1. 2x + 5 = 18 2. 4 -1=0
w
(w 2 - 1)(w2 +1) = 0
x = 13/2 (w2 - 1)(w2 - [-1]) = 0
(w - 1)(w + 1)(w - i)(w + i)
Wednesday, March 18, 2009

7. Warm-up:
Solve for all possible solutions (real and imaginary).
1. 2x + 5 = 18 2. 4 -1=0
w
(w 2 - 1)(w2 +1) = 0
x = 13/2 (w2 - 1)(w2 - [-1]) = 0
(w - 1)(w + 1)(w - i)(w + i)
w = 1, -1, i, -i
Wednesday, March 18, 2009

8. Fundamental Theorem
of Algebra
Wednesday, March 18, 2009

9. Fundamental Theorem
of Algebra
Every polynomial equation P(x) = 0 of any degree
with complex coefficients has at least one complex
number solution.
Wednesday, March 18, 2009

10. Double Root
Wednesday, March 18, 2009

11. Double Root
In a quadratic, when the discriminant equals 0,

there will be two roots that have the same value.
Wednesday, March 18, 2009

12. Double Root
In a quadratic, when the discriminant equals 0,

there will be two roots that have the same value.
When any root appears twice.

Wednesday, March 18, 2009

13. Multiplicity of a Root
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14. Multiplicity of a Root
The highest power of (x - r) of a polynomial when r is
a root.
Wednesday, March 18, 2009

15. The Number of Roots
of a Polynomial
Equation Theorem
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16. The Number of Roots
of a Polynomial
Equation Theorem
Every polynomial of degree n has exactly n roots
(including multiplicity).
Wednesday, March 18, 2009

17. Example 1: How many
roots does each
equation have?
4 2
a. x15 +1=0 b. 2x − 3x + π = 0
Wednesday, March 18, 2009

18. Example 1: How many
roots does each
equation have?
4 2
a. x15 +1=0 b. 2x − 3x + π = 0
15
Wednesday, March 18, 2009

19. Example 1: How many
roots does each
equation have?
4 2
a. x15 +1=0 b. 2x − 3x + π = 0
4
15
Wednesday, March 18, 2009

20. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
Wednesday, March 18, 2009

21. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2:
Wednesday, March 18, 2009

22. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2
Wednesday, March 18, 2009

23. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2
q = factors of 1:
Wednesday, March 18, 2009

24. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2
q = factors of 1: ±1
Wednesday, March 18, 2009

25. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2 p
= ±1,±2
q
q = factors of 1: ±1
Wednesday, March 18, 2009

26. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2 p
= ±1,±2
q
q = factors of 1: ±1
Wednesday, March 18, 2009

27. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2 p
= ±1,±2
q
q = factors of 1: ±1
Wednesday, March 18, 2009

28. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2 p
= ±1,±2
q
q = factors of 1: ±1
Wednesday, March 18, 2009

29. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2 p
= ±1,±2
q
q = factors of 1: ±1
Wednesday, March 18, 2009

30. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2: ±1,±2 p
= ±1,±2
q
q = factors of 1: ±1
Wait a minute! Why isn’t the zero
of this function either 1 or 2?
Wednesday, March 18, 2009

31. The rational-root theorem only tells us where to look for
rational roots. What does this tell us about the roots of
our function?
Wednesday, March 18, 2009

32. The rational-root theorem only tells us where to look for
rational roots. What does this tell us about the roots of
our function?
So let’s graph on the following window:
-5 ≤ x ≤ 10 and -110 ≤ y ≤ 40
Wednesday, March 18, 2009

33. The rational-root theorem only tells us where to look for
rational roots. What does this tell us about the roots of
our function?
So let’s graph on the following window:
-5 ≤ x ≤ 10 and -110 ≤ y ≤ 40
Wednesday, March 18, 2009

34. The rational-root theorem only tells us where to look for
rational roots. What does this tell us about the roots of
our function?
So let’s graph on the following window:
-5 ≤ x ≤ 10 and -110 ≤ y ≤ 40
Wednesday, March 18, 2009

35. The rational-root theorem only tells us where to look for
rational roots. What does this tell us about the roots of
our function?
So let’s graph on the following window:
-5 ≤ x ≤ 10 and -110 ≤ y ≤ 40
Wednesday, March 18, 2009

36. The rational-root theorem only tells us where to look for
rational roots. What does this tell us about the roots of
our function?
So let’s graph on the following window:
-5 ≤ x ≤ 10 and -110 ≤ y ≤ 40
With this being a quartic (4th degree), we also know that
there should be 3 changes in the curvature. Are there?
Wednesday, March 18, 2009

37. The rational-root theorem only tells us where to look for
rational roots. What does this tell us about the roots of
our function?
So let’s graph on the following window:
-5 ≤ x ≤ 10 and -110 ≤ y ≤ 40
With this being a quartic (4th degree), we also know that
there should be 3 changes in the curvature. Are there?
There are 2 real solutions.
Wednesday, March 18, 2009

38. b. How many roots (real or complex) are there?
Wednesday, March 18, 2009

39. b. How many roots (real or complex) are there?
x4 - 6x3 + 2x2 - 3x + 2 = 0
Wednesday, March 18, 2009

40. b. How many roots (real or complex) are there?
x4 - 6x3 + 2x2 - 3x + 2 = 0
Wednesday, March 18, 2009

41. b. How many roots (real or complex) are there?
x4 - 6x3 + 2x2 - 3x + 2 = 0
4
Wednesday, March 18, 2009

42. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
Wednesday, March 18, 2009

43. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2
Wednesday, March 18, 2009

44. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2
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45. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x
Wednesday, March 18, 2009

46. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x + 25) = 0
Wednesday, March 18, 2009

47. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x + 25) = 0
x2(x + 5)2 = 0
Wednesday, March 18, 2009

48. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x + 25) = 0
x2(x + 5)2 = 0
x = 0, -5
Wednesday, March 18, 2009

49. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x + 25) = 0
x2(x + 5)2 = 0
x = 0, -5
What can we say about these roots?
Wednesday, March 18, 2009

50. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x + 25) = 0
x2(x + 5)2 = 0
x = 0, -5
What can we say about these roots?
They are each double roots.
Wednesday, March 18, 2009

51. Homework:
Wednesday, March 18, 2009

52. Homework:
p. 721 #1-21
Wednesday, March 18, 2009

53. Homework:
p. 721 #1-21
“Winning isn’t everything, but wanting to win is.”
-Vince Lombardi
Wednesday, March 18, 2009