1. Apply the Fundamental Theorem of Algebra
Section 5-7
Apply the Fundamental Theorem of Algebra

2. Example 1 How many solutions does each equation have?
a.) x4 + 8x2 – 5x + 2
b.) x3 + x2 – 3x – 3
Four
Three
= 5g3(g + 4)(g – 4)

3. Example 2 Find all real zeros of f(x) = x5 – 4x3 + x2 – 4
Step 1: List possible rational zeros.
± 1, ± 2, ± 4 ±1
Step 2: Test the zeros using synthetic division.
Test x = -1 -1
– –4 -1 1 3 -4 4 1 -1 -3 4 -4
x4 – x3 – 3x2 + 4x – 4

4. Example 2 Find all real zeros of f(x) = x4 – x3 – 3x2 + 4x – 4
Step 3: List possible rational zeros.
± 1, ± 2, ± 4 ±1
Step 4: Test the zeros using synthetic division.
Test x = -2 -2
– –4 -2 6 -6 4 1 -3 3 -2
x3 – 3x2 + 3x – 2
= 5g3(g + 4)(g – 4)

5. Example 2 - continued f(x) = x3 – 3x2 + 3x – 2
Step 5: Test the zeros using synthetic division.
Test x = 2 2
1 – 2 2 -2 1
- 1 1
x2 – x + 1

6. Example 2 - continued
Step 6: Find the remaining zeros by solving x2 – x + 1 = 0.
Use the quadratic formula:
x = -b ± √b2 – 4ac 2a
x = 1 ± i√3 2
x = 1 ± √(-1)2 – 4(1)(1)
2(1)
x = 1 ± √1 – (4) 2
x = 1 ± √-3 2
The real zeros of f are -1, 2, -2, 1 + i√3, 1 – i√3
= 5g3(g + 4)(g – 4)

7. Homework
Section 5-7
Pages 383 – 384
3 – 6,
9, 10
13 – 16

8. The Fundamental Theorem of Algebra
Theorem: If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one solution in the set of complex numbers.
Corollary: If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has exactly n solutions provided each solution repeated twice is counted as 2 solutions, each solution repeated three times is counted as 3 solutions and so on.