1. Lesson: _____ Section 2.5 The Fundamental Theorem of Algebra
Counting complex & repeated solutions, an nth degree polynomial function has exactly n zeros. (*double roots count twice!)
If a + bi is a zero, then a-bi is also a zero.
 Complex zeros always travel in conjugate pairs.

2. Express as a product of linear factors and find all zeros.
f(x) = x5 + x3 + 2x2 – 12x + 8

4. Ex. Given that (-2 + i) is a zero of f(x), find the remaining zeros.
f(x) = x4 + 4x3 + 12x2 + 28x + 35

5. Distributing this would produce x2 – 3x – xi + 3i
No need to write this down. Just follow along…
Why do the complex roots have to travel in pairs?
What if x = 3 and x = i were the only roots?
Then (x – 3) and (x – i) would be factors.
(x – 3)(x – i)
Distributing this would produce x2 – 3x – xi + 3i
…which is not a polynomial! (all polynomials must have real coefficients)
The only way for an imaginary root to be acceptable is if somehow the i’s disappeared. This only occurs when you multiply conjugates.