2.7 Apply the Fundamental Theorem of Algebra
Transcript of 2.7 Apply the Fundamental Theorem of Algebra
2.7 Apply the Fundamental Theorem of Algebra
Fundamental Theorem of Algebra
Every non-constant single-variable polynomial with complex coefficients has
at least one complex root.
Theorem
• Every polynomial equation with complex coefficients and positive degree n has exactly n roots.
• Textbook p. 96 # 2 and 4
Conjugate Root Theorem
• If a polynomial equation with real coefficients has a + bi as a root (a and b real, b not = 0), then a - bi is also a root.
• If P(x) has 3 – 4i as a root, what is also a root?
3 + 4i• If P(x) has -2 + i as a root, what is also a root?
-2 - i
Find a cubic function with integral coefficients that has 2 and 3 – i as roots.
• How many roots does the equation have?
3
• What are two of the roots?
2 and 3 - i
• What is the third root?
3 + i
• What are the factors?
(x - 2)(x - (3 - i))(x – (3 + i))
So what's the function?
• f(x) = (x - 2)(x - (3 - i))(x – (3 + i))
• Simplify the function.
• f(x) = x³ - 8x² + 22x - 20
Solve x⁴ - 12x – 5 = 0 given that -1 + 2i is a root.
• How many roots does the equation have?4
• What is one root? -1 + 2i
• Then what is one factor?(x – (-1 + 2i)) = (x + 1 – 2i)
• What is another root?-1 – 2i
• Then what is another factor?(x – (-1 + 2i)) = (x + 1 – 2i)
• How can we find the other factors?
x⁴ - 12x – 5 = 0
• (x – (-1 + 2i)) (x – (-1 – 2i)) = x² + 2x + 5
Use long division to find the other factors.
(x² + 2x + 5)(x² - 2x - 1)
In this case there are two factors.
Classwork
• Textbook p. 96 # 16
Irrational Conjugates Theorem
• If a + √b is a zero of P(x) ,
then a - √b is also a zero of P(x).
• If P(x) has 3 – √4 as a root, what is also a root?
3 + √4• If P(x) has -2 + √3 as a root, what is also a root?
-2 - √3
Find a cubic function with integral coefficients that has 2 and 3 – √2 as roots.
• How many roots does the equation have?
3
• What are two of the roots?
2 and 3 - √2
• What is the third root?
3 + √2
• What are the factors?
(x - 2)(x - (3 - √2))(x – (3 + √2))
(x – 2)(x - 3 + √2)(x – 3 - √2)
Descartes' Rule of Signs
• Let P(x) be a simplified polynomial with real coefficients and terms arranged in decreasing degree of x.
1. The number of positive real roots of P(x) = 0 equals the number of variations of sign of P(x) or is fewer than this number by an even integer.
2. The number of negative real roots of P(x) = 0 equals the number of variations of sign of P(-x) or is fewer than this number by an even integer.
P(x) = x⁵ + x⁴ - 3x² + 4x + 6• How many variations of sign does P(x) have?
2: So the number of positive real roots is 2 or 0.
• How many variations of sign does P(-x) have?
• P(-x) = (-x)⁵ + (-x)⁴ - 3(-x)² + 4(-x) + 6
= -x⁵ + x⁴ - 3x² - 4x + 6
3: So the number of negative real roots is 3 or 1.
• How many roots does P(x) have?
5 : They can be positive real roots , negative real roots or imaginary roots.
P(x) = x⁵ + x⁴ - 3x² + 4x + 6 has 5 roots.
• P(x) = x⁵ + x⁴ - 3x² + 4x + 6 has 2 variations of sign. The number of positive real roots is 2 or 0.
• P(-x) = (-x)⁵ + (-x)⁴ - 3(-x)² + 4(-x) + 6
= -x⁵ + x⁴ - 3x² - 4x + 6 has 3 variations of sign. The number of negative real roots is 3 or 1.
• How many imaginary roots?Total number of
Roots# Positive Real
Roots# Negative Real
Roots# Imaginary
Roots
5 2
5 2
5 0
5 0
Total number of Roots
# Positive Real Roots
# Negative Real Roots
# Imaginary Roots
5 2 3
5 2 1
5 0 3
5 0 1
Total number of Roots
# Positive Real Roots
# Negative Real Roots
# Imaginary Roots
5 2 3 0
5 2 1 2
5 0 3 2
5 0 1 4
• Use Decartes' Rule of Signs to fill in the table.
Total number of Roots
# Positive Real Roots
# Negative Real Roots
# Imaginary Roots
Total number of Roots
# Positive Real Roots
# Negative Real Roots
# Imaginary Roots
5 3 2 0
5 3 0 2
5 1 2 2
5 1 0 4
Homework
• Textbook page 96 # 1 – 17 odd
page 97 # 15 - 18