January 19, 2012 At the end of today, you will be able to find ALL zeros of a polynomial. Warm-up: Check HW 2.4 Pg. 167 #20-21, 27-29, 46- 49, 55, 65,

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January 19, 2012 At the end of today, you will be able to find ALL zeros of a polynomial. Warm-up: Check HW 2.4 Pg. 167 #20-21, 27-29, , 55, 65, – 11i47. 8/41 + (10/41)i i√248. 5/2 + (5/2)i i49. 4/5 + (3/5)i – 22i55. -1/2 – (5/2)i i65. 1 ± i 46. 7i ± (1/2)i

Lesson 2.5 Zeros of Polynomial Functions The Fundamental Theorem of Algebra tells you how many zeros or factors of a polynomial exist. If f(x) is a polynomial of degree n, where n > 0, then f has at least one zero in the complex number system. For example: a)f(x) = x – 2, b)f(x) = x 2 – 6x + 9 = (x – 3)(x – 3), c)f(x) = x 3 + 4x = x(x 2 + 4) = x(x -2i)(x +2i), has 2 zeros: x = 3 (multiplicity of 2) has exactly 1 zero: x = 2 has 3 zeros x = 0, x = 2i, x = -2i

Using the Rational Zero Test to find real zeros. The Rational Zero Test helps you find the rational zeros of polynomials. Rational zero =, where p and q have no common factors other than 1, and p = a factor of the constant term a 0 q = a factor of the leading coefficent a n for f(x) = a n x n + a n-1 x n-1 +…+ a 2 x 2 + a 1 x + a 0 Example 1: Use the Rational Zero Test to find possible zeros for f(x) = 2x 3 + 3x 2 – 8x + 3

Finding the zeros After you use the Rational Zero Test, use synthetic division to find zeros. Pick one of the zeros from the test and hope you get a remainder of 0! (1 is the easiest) Example 1: Use the Rational Zero Test to find possible zeros for f(x) = 2x 3 + 3x 2 – 8x = 2x 2 + 5x – 3 So your factors are: (x – 1)( 2x 2 + 5x – 3) (x – 1)( 2x – 1)(x + 3) Zeros = 1, 1/2, -3

Use the Rational Zero Test and Synthetic/Long Division to find the factors and zeros. Example 2: Find all the zeros of the function and write the polynomial as a product of linear factors. f(x) = x 5 + x 3 + 2x 2 – 12x + 8 Try x 4 + x 3 + 2x 2 + 4x – 8 Try another zero: x 3 – x 2 + 4x – 4 So far we have (x – 1)(x + 2)(x 3 – x 2 + 4x – 4)

NOW FACTOR x 2 (x – 1) + 4(x – 1) (x – 1)(x 2 + 4) x = 1 and x = ±2i Finally! Factors: (x – 1)(x + 2)(x + 1)( x + 2i)(x – 2i) Zeros: x = 1, -2, -1, 2i, -2i (x – 1)(x + 2)(x 3 – x 2 + 4x – 4) *Complex Zeros occur in Conjugate Pairs! (a + bi)(a – bi)