This document discusses solving polynomial equations. It begins by working through examples of solving quadratic equations. It then introduces concepts like the Fundamental Theorem of Algebra, which states that every polynomial equation has at least one complex number solution. It discusses double roots and the multiplicity of roots. Finally, it works through examples of determining the number of real solutions a polynomial equation would have based on its degree.

1. Section 11-8
Solving All Polynomial Equations
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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
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8. Fundamental Theorem
of Algebra
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9. Fundamental Theorem
of Algebra
Every polynomial equation P(x) = 0 of any degree
with complex coefficients has at least one complex
number solution.
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10. Double Root
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11. Double Root
In a quadratic, when the discriminant equals 0,

there will be two roots that have the same value.
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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.

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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.
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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).
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17. Example 1: How many
roots does each
equation have?
4 2
a. x15 +1=0 b. 2x − 3x + π = 0
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18. Example 1: How many
roots does each
equation have?
4 2
a. x15 +1=0 b. 2x − 3x + π = 0
15
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19. Example 1: How many
roots does each
equation have?
4 2
a. x15 +1=0 b. 2x − 3x + π = 0
4
15
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20. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
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21. Example 2:
Consider the equation x4 - 6x3 + 2x2 - 3x + 2 = 0.
a. How many real solutions are there?
p = factors of 2:
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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
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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:
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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
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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?
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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?
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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.
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38. b. How many roots (real or complex) are there?
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39. b. How many roots (real or complex) are there?
x4 - 6x3 + 2x2 - 3x + 2 = 0
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40. b. How many roots (real or complex) are there?
x4 - 6x3 + 2x2 - 3x + 2 = 0
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41. b. How many roots (real or complex) are there?
x4 - 6x3 + 2x2 - 3x + 2 = 0
4
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42. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
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43. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2
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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
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46. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x + 25) = 0
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47. Example 3:
Find all roots of 4 + 10x3 + 25x2 =0
x
x2(x2 + 10x + 25) = 0
x2(x + 5)2 = 0
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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.
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51. Homework:
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52. Homework:
p. 721 #1-21
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53. Homework:
p. 721 #1-21
“Winning isn’t everything, but wanting to win is.”
-Vince Lombardi
Wednesday, March 18, 2009