ALGEBRA II HONORS/GIFTED @ ALGEBRA II HONORS/GIFTED - SECTIONS 5-5 and 5-6 (The Fundamental Theorem of Algebra), Descrates' Rule of Signs ALGEBRA II HONORS/GIFTED.

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ALGEBRA II HONORS/GIFTED @ ALGEBRA II HONORS/GIFTED - SECTIONS 5-5 and 5-6 (The Fundamental Theorem of Algebra), Descrates' Rule of Signs ALGEBRA II HONORS/GIFTED @ SECTIONS 5-5 and 5-6 : THE FUNDAMENTAL THEOREM OF ALGEBRA DECSARTES’ RULE OF SIGNS

1) Factor completely. 2) Now, find the zeros of each polynomial. a) 2x2 – 5x - 12 ANSWER : (2x + 3)(x – 4) 3) What do you notice about each of the zeros in relation to the lead coefficient and the constant terms of each polynomial? b) x2 - 36 ANSWER : (x + 6)(x – 6) c) x3 – 2x2 – 4x + 8 ANSWER : (x + 2)(x – 2)2

REMAINDER THEOREM : means you will always get a remainder, even if it is zero, when you divide a polynomial by another polynomial. RATIONAL ROOT THEOREM : ax – b is a factor of P(x) iff CONJUGATE ROOT THEOREM : If P(x) is a polynomial with rational coefficients, then the following must be true : -If is a root of P(x), then is also a root. -If a + bi is a root of P(x), then a – bi is also a root.

4) Factor x3 – 6x2 – x + 30 It looks like we need another method for factoring. Use the Rational Root Theorem to determine the possible zeros of the polynomial. The possible rational roots are Test each of the prospective roots to determine what works. We can determine that -2 is a root. How do we figure out the remaining roots?

Use synthetic division to determine the other roots. Divide -2 into x3 – 6x2 – x + 30. We get x2 – 8x + 15. Factor it and get the other roots. a) What are the roots for x3 – 6x2 – x + 30? ANSWER : -2, 3, 5 b) What are the factors? ANSWER : (x + 2)(x – 3)(x – 5)

5) Use the Rational Root Theorem to list all possible rational roots for each equation. Then find any actual rational roots. a) x4 – 7x3 + 18x2 – 20x + 8 = 0 ANSWERS : 1 and 2 b) x3 – 5x2 - 2x + 10 = 0 ANSWER : 5 c) 6x3 – 31x2 + 25x + 12 = 0 ANSWERS :

6) A polynomial function P(x) with rational coefficients has the given roots. Find two additional roots of P(x) = 0. 6 + 5i and ANSWERS : 6 – 5i and 7) Write a polynomial function with rational coefficients so that P(x) = 0 has the given roots. a) -5 and 7 c) 1, 3, and 1 - 2i ANSWER : x2 -2x – 35 = 0 x4 – 6x3 + 16x2 - 26x + 15 = 0 b) 7 + 4i ANSWER : x2 – 14x + 65 = 0

DESCARTES’ RULE OF SIGNS : 1. The number of positive zeros of a polynomial is less than or equal to the number of sign reversals in P(x). 2. The number of negative zeros is less than or equal to the number of sign reversals in P(-x). 3. In both cases, the number of zeros has the same parity (even or odd, that is, decreases by 2 until you reach 0 or 1) as the number of sign reversals. Rene Descartes 1596-1650

8) Use Descartes’ Rule of Signs to determine the possible zeros for P(x) = x5 – 7x4 – 5x3 + 11x2 – 2x + 5 # of Sign Changes: P(-x)= Solution: +R -R I = Total

Use your to graph P(x) = x5 – 7x4 – 5x3 + 11x2 – 2x + 5 and determine the correct arrangements of +R, -R, and I zeroes.

9) Use Descartes’ Rule of Signs to determine the possible zeros. a) P(x) = 2x7 - 3x5 – x4 – x3+ x – 16 b) P(x) = 2x4 – 3x3 – 2x2 + x + 4