1. 7.4 The Fundamental Theorem of Algebra

2. Fundamental Theorem of Algebra: Every polynomial function of positive degree with complex coefficients has at least one complex zero Thm: If a polynomial is of degree n, then it has n linear factors and n zeros. *Note: these zeros may have duplicates these zeros may or may not be real Reminders: *Algebra with i’s* Addition: (a + bi) + (c + di) = (a + c) + (b + d)i ex: (6 + 2i) + (4 – 5i) Subtraction: (a + bi) – (c + di) = (a – c) + (b – d)i ex: (2 + i) – (6 – i)  We can get answers with i in them (multiplicity) = 10 – 3i = –4 + 2i

3. Reminders: *Algebra with i’s* (cont…) Multiplication: (a + bi)(c + di) = (ac – bd) + (ad + bc)i ex: (6 + i)(3 + 2i) Division: Complex Conjugate Theorem: If a complex number a + bi is a zero of a polynomial, then so is its conjugate a – bi Conjugate Radical Theorem: If is a zero of a polynomial, then so is its conjugate *Both complex & radical roots come in pairs!* = 18 + 12i + 3i + 2i 2 = 16 + 15i w/ real coeff w/ rational coeff

4. Reminder: to get a quadratic from 2 roots we can use x 2 – (sum)x + product Ex: roots are 2 and 6 Ex: roots are 2 + 3i and 2 – 3i Ex: roots are and (+): 2 + 6 = 8 (×): 2(6) = 12 x 2 – 8x + 12 (+): (2 + 3i) + (2 – 3i) (×): (2 + 3i)(2 – 3i) = 4 = 4 – 9i 2 = 4 + 9= 13 x 2 – 4x + 13 (+): (×): = 6 = 9 – 6= 3 x 2 – 6x + 3

5. Now let’s put all this new info and old algebra skills together to determine roots of polynomials & write equations of polynomials! Ex 1) Determine the zeros of *because we might have to do multiple synthetic substitutions, we will stack them all together & do the “middle row” in our head* now do S.D. on depressed equation new coefficients!! Now, depressed eqtn is a quadratic… solve it!! 1 1 1 –1 1 –2 1 –1 2 11 12 02 2 0 –2 14920 1010

6. –1 + i1 2 0 –4 –4 ↓ 11 + i –2 0 –1 + i –2 – 2i 2 – 2i –1 – i ↓ 10 2 + 2i–1 – i 0 Now do S.D. again on the result with –1 – i Ex 2) Determine all the zeros given 1 zero. If –1 + i is a zero, so is… Two Methods to solve! (1) Synthetically substitute each to get the depressed eqtn –1 – i 4 –2 0 Solve quadratic 

7. You must get zero or you did something wrong Now solve x 2 – 2 = 0 OR (2) Find quadratic with 2 complex roots, do long division & then depressed equation zeros: –1 + i, –1 – i (+): –1 + i + –1 – i (×): (–1 + i)(–1 – i) = –2 = 1 – i 2 = 1 + 1= 2 x 2 + 2x + 2 * Choose whichever method you are most comfortable with!

8. Ex 3) Determine the polynomial of lowest degree with real coefficients that has and –2 as zeros Now, if –2 is a zero, the factor is (x + 2) So … the polynomial is Let’s find quadratic with & as zeros first If is a zero, so is… (+): (×): = 4 = 4 – 5i 2 = 4 + 5 = 9 x 2 – 4x + 9 with –2 also, lowest degree is 3

9. Homework #704 Pg 353 #1, 3, 5, 7, 11, 15, 17, 18, 21–29 odd, 34, 37, 41, 43, 49