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)

HW Answers: Pg 258 (7, 11, 18, 20, 21, 32, 34, 51-58 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) 16 41 + 20 41 𝑖 54) 3 2 + 3 2 𝑖 55) 3 5 + 4 5 𝑖 56) 22 5 + 9 5 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) -4 + 2i 85) i 86) 1 8 i

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

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.

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

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.

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

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

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.

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

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.

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 + 5 )(x - 5 )(x2 + 4) c. x4 – x2 – 20 = (x + 5 )(x - 5 )(x – 2i)(x + 2i)

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 -27+15i -45-6i 18 3i 3 5 25 45 -18 3 5+9i -2+15i -6i 0 3 5 -2 0 -3i -9i -15i 6i 0 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.

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)].

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

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)