1. Warm-up: Find all real solutions of the equation X4 – 3x2 + 2 = 0
HW: pg 266 (13-27 odd, 35, 37, 39, 44, 49, 51, 53)

2. HW Answers: Pg 258 (7, 11, 18, 20, 21, 32, 34, all, 66, 68, 78, 79, 80, 85, 86)
7) 2 – 3i 3
11) -1 – 6i
18) 8 + 4i
20) -3 – 11i
21) 3 – 3i 2
32) 6 – 22i
34) 32 – 72i
51) -6i
52) 5i
53) 𝑖
54) 𝑖
55) 𝑖
56) 𝑖
57) – 7 – 6i
58) 10 – 4i
66) -3 ± i
68) 1 3 ±2𝑖
78) a.1 b.i c.-1 d.-i
79) – 1 + 6i
80) i
85) i
86) 1 8 i

3. Imaginary & Complex Numbers
Objective:
Identify, add, subtract, multiply, and divide imaginary and complex numbers
Finding complex solutions of a Quadratic Equation

4. The Fundamental Theorem Of Algebra
Every Polynomial Equation with a degree higher than zero has at least one root in the set of Complex Numbers.

5. Linear Factorization Theorem
If f(x) is a polynomial of degree n
where n > 0, then f has precisely n linear factors
where are complex numbers

6. Real/Imaginary Roots
Just because a polynomial has ‘n’ complex roots doesn’t mean that they are all Real!
In this example, however, the degree is still n = 3, but there is only one Real x-intercept or root at x = -1, the other 2 roots must have imaginary components.

7. Zeros of Polynomial Functions
Give the degree of the polynomial, tell how many zeros there are, and find all the zeros
One zero: x = 2
1st degree
= (x – 3)(x – 3)
Two zeros with dual multiplicity known as repeated zeros: x = 3 and x = 3
2nd degree
= x(x2 + 4) = x(x – 2i)(x + 2i)
Three zeros: x = 0, x = 2i and x = -2i
3rd degree
= (x – 1)(x + 1)(x – i)(x + i)
Four zeros: x = 1, x = -1, x = i and x = -i
4th degree

8. Example: find ALL the zeros of the function.1 1 1 2
-12 8 1 1 2 4 -8 1 1 2 4 -8 1 1 1 2 4 -8 1 2 4 8 1 2 4 8 -2 1 2 4 8 -2 -8 1 4

9. Complex Conjugates Theorem
Roots/Zeros that are not Real are Complex with an Imaginary component. Complex roots with Imaginary components always exist in Conjugate Pairs.
If a + bi (b ≠ 0) is a zero of a polynomial function, then its Conjugate, a - bi, is also a zero of the function.

10. f(x)= (x – 1)(x – 1)(x – 3i)(x + 3i) f(x)= (x2 – 2x + 1)(x2 + 9)
Finding a Polynomial with Given Zeros: Write a polynomial function of least degree that has real coefficients, that has 1, 1, and 3i as zeros.
f(x)= (x – 1)(x – 1)(x – 3i)(x + 3i)
f(x)= (x2 – 2x + 1)(x2 + 9)
f(x)= x4 – 2x3 + 10x2 – 18x + 9

11. Factors of a Polynomial with Real Coefficients
Definition: A quadratic with no real zeros
is irreducible over the reals.
Every polynomial function with real coefficients
can be written as a product of linear factors
and irreducible quadratic factors, each with
real coefficients.

12. Factoring a Polynomial
Write the polynomial f(x) = x4 – x2 – 20
As the product of factors that are irreducible over the rationals
As the product of linear factors and quadratic factors that are irreducible over the reals
In completely factored form
a. x4 – x2 – 20 = (x2 – 5)(x2 + 4)
b. x4 – x2 – 20 = (x )(x )(x2 + 4)
c. x4 – x2 – 20 = (x )(x )(x – 2i)(x + 2i)

13. Use the given root to find the remaining roots of the function
Since 3i is a root we also know that its conjugate -3i is also a root. Let’s use synthetic division and reduce our polynomial by these roots then.
9i i i 3i
i i i
-3i
-9i i i
So the roots of this function are 3i, -3i, 1/3, and -2
Put variables in here & set to 0 and factor or formula this to get remaining roots.

14. Find Roots/Zeros of a Polynomial
Ex: Find all the roots of
If one root is 4 - i.
Because of the Complex Conjugate Theorem, we know that another root must be 4 + i.
Can the third root also be imaginary?
If one root is 4 - i, then one factor is [x - (4 - i)], and
Another root is 4 + i, & another factor is [x - (4 + i)].

15. Multiply these factors:
Use this trinomial (a factor of the original polynomial) to find another factor of the polynomial by dividing.
The third root is x = -3

16. Sneedlegrit: Find all the zeros of given that 1+3i is a
is a zero of f.
HW: pg 266 (13-27 odd, 35, 37, 39, 44, 49, 51, 53)