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7.4 The Fundamental Theorem of Algebra

7.4 The Fundamental Theorem of Algebra. 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.

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7.4 The Fundamental Theorem of Algebra

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  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) (multiplicity)  We can get answers with i in them • = 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 + 2i2 • = 16 + 15i w/ real coeff w/ rational coeff

  4. Reminder: to get a quadratic from 2 roots we can use x2 – (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 • x2 – 8x + 12 • (+): (2 + 3i)+ (2 – 3i) • (×): (2 + 3i)(2 – 3i) = 4 • x2 – 4x + 13 • = 4 + 9 • = 13 • = 4 – 9i2 • (+): • (×): = 6 • x2 – 6x + 3 • = 9 – 6 • = 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* 1 1 –1 1 –2 now do S.D. on depressed equation new coefficients!! 1 1 2 1 2 0 –1 1 1 0 2 2 1 4 9 20 –2 1 0 1 0 Now, depressed eqtn is a quadratic… solve it!!

  6. 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 –1 + i 1 2 0 –4 –4 ↓ 4 –1 + i –2 2 – 2i –2 – 2i –2 –1 – i 1 1 + i 0 ↓ Now do S.D. again on the result with –1 – i –1 – i 2 + 2i 0 0 1 0 –2 Solve quadratic 

  7. You must get zero or you did something wrong Now solve x2 – 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 • x2 + 2x + 2 • = 1 + 1 • = 2 • = 1 –i2 * 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 If is a zero, so is… • with –2 also, lowest degree is 3 Let’s find quadratic with & as zeros first • (+): • (×): = 4 • x2 – 4x + 9 • = 4 – 5i2 • = 4 + 5 = 9

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

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