1. Algebra 2

2. Solve for x

3. Algebra 2 (KEEP IN MIND THAT A COMPLEX NUMBER CAN BE REAL IF THE IMAGINARY PART OF THE COMPLEX ROOT IS ZERO!) Lesson 6-6 The Fundamental Theorem of Algebra Additional Examples

4. Algebra 2 For the equation x 4 – 3x 3 + 4x + 1 = 0, find the number of complex roots, the possible number of real roots, and the possible rational roots. Lesson 6-6 The Fundamental Theorem of Algebra By the corollary to the Fundamental Theorem of Algebra, x 4 – 3x 3 + 4x + 1 = 0 has four complex roots. By the Imaginary Root Theorem, the equation has either no imaginary roots, two imaginary roots (one conjugate pair), or four imaginary roots (two conjugate pairs). So the equation has either zero real roots, two real roots, or four real roots. By the Rational Root Theorem, the possible rational roots of the equation are ±1. Additional Examples

5. Algebra 2 Find the number of complex zeros of ƒ(x) = x 5 + 3x 4 – x – 3. Find all the zeros. Lesson 6-6 The Fundamental Theorem of Algebra By the corollary to the Fundamental Theorem of Algebra, there are five complex zeros. Additional Examples Step 1: Find a rational root from the possible roots of ±1 and ±3.

6. Algebra 2 Lesson 6-6 The Fundamental Theorem of Algebra Additional Examples ƒ(x) = x 5 + 3x 4 – x – 3

7. Algebra 2 (continued) Lesson 6-6 The Fundamental Theorem of Algebra Step 3: Factor the expression x 4 – 1. x 4 – 1 = (x 2 – 1)(x 2 + 1) Difference of Two Squares = (x – 1)(x + 1)(x 2 + 1) Factor x 2 – 1. So 1 and –1 are also roots. Additional Examples

8. Algebra 2 Step 3: Solve x 2 + 1 = 0. The polynomial function ƒ(x) = x 5 + 3x 4 – x – 3 has three real zeros of x = –3, x = 1, and x = –1, and two complex zeros of x = i, and x = –i. Lesson 6-6 The Fundamental Theorem of Algebra Additional Examples x 2 + 1= 0 x 2 = –1 x= ± i So i and –i are also roots.

9. Algebra 2 Find the number of complex zeros of ƒ(x) = x 5 + 3x 4 – x – 3. Find all the zeros. Lesson 6-6 The Fundamental Theorem of Algebra By the corollary to the Fundamental Theorem of Algebra, there are five complex zeros. You can use synthetic division to find a rational zero. Step 1: Find a rational root from the possible roots of ±1 & ±3. ƒ(-3) = (-3) 5 + 3(-3) 4 – (-3) – 3 ƒ(-3) = (-3) 5 + (-3) 5 +3 – 3 ƒ(-3) = 0 So –3 is one of the roots. Additional Examples Step 2: Factor the expression x 4 – 1. x 4 – 1 = (x 2 – 1)(x 2 + 1) Factor x 2 – 1. = (x – 1)(x + 1)(x 2 + 1) So 1 and –1 are also roots. Step 3: Solve x 2 + 1 = 0. x 2 + 1= 0 x 2 = –1 x = ± i So i and –i are also roots. The polynomial function ƒ(x) = x 5 + 3x 4 – x – 3 has three real zeros of x = –3, x = 1, and x = –1, and two complex zeros of x = i, and x = –i.