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