1. Section 2.5 Part Two Other Tests for Zeros Descartes’s Rule of Signs Upper and Lower Bounds

2. Descartes’s Rule of Signs Let f(x) be a polynomial with real coefficients and a o ≠ 0. The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integer. The number of negative real zeros of f is either equal to the number of variations in the sign of f(-x) of less than that number by an even integer.

3. Apply Descarte’s Rule of Signs Consider f(x) = x 2 + 8x + 15 Since there are no variations in sign of f(x) there are no positive roots. Since f(-x) = (-x) 2 + 8(-x) + 15 = x 2 – 8x + 15 has two variations, f(x) may have two or zero negative roots. x = {-3. -5}

4. Variation in Sign A variation in sign means that when the polynomial is written in standard form that one term has a different sign than the next. T(x) = x 5 + 9x 4 + 19x 3 – 21x 2 – 92x – 60 For Variations of T(x) just look at the signs T(x) has one variation in signs For Variations of T(-x) you must change the sign of terms with odd numbered exponents T(-x) = -x 5 + 9x 4 – 19x 3 – 21x 2 + 92x – 60 T(-x) has four variation in signs

5. Find the Zeros of T(x) T(x) = x 5 + 9x 4 + 19x 3 – 21x 2 – 92x – 60 60  + 1, + 2, + 3, + 4, + 5, + 6, + 10, + 12, + 15, + 20, + 30, + 60 Since there is only one sign variation for T(x) there is at most, one positive root SYN Program

6. Find the Zeros of T(x) T(x) = x 5 + 9x 4 + 19x 3 – 21x 2 – 92x – 60 T(x) = (x – 2)(x 4 + 11x 3 + 41x 2 + 61x + 30) 1919-21-92 +2 1 2 11 22 41 82 61 122 30 -60 60 0 Now that we have found the positive root we know that any other real roots must be negative.

7. Find the Zeros of T(x) T(x) = x 5 + 9x 4 + 19x 3 – 21x 2 – 92x – 60 T(x) = (x – 2)(x 4 + 11x 3 + 41x 2 + 61x + 30) 30  -1, -2, -3, -4, -5, -6, -10,-15, - 30 T(x) = (x – 2)(x + 1)(x 3 + 10x 2 + 31x + 30) 111416130 1 10 -10 31 -31 30 -30 0

8. Find the Zeros of T(x) T(x) = x 5 + 9x 4 + 19x 3 – 21x 2 – 92x – 60 T(x) = (x – 2)(x 4 + 11x 3 + 41x 2 + 61x + 30) T(x) = (x – 2)(x + 1)(x 3 + 10x 2 + 31x + 30) 30  -1, -2, -3, -4, -5, -6, -10,-15, - 30 T(x) = (x – 2)(x + 1)(x + 2)(x 2 + 8x + 15) 1103130 -2 1 8 -16 15 -30 0

9. Find the Zeros of T(x) T(x) = x 5 + 9x 4 + 19x 3 – 21x 2 – 92x – 60 T(x) = (x – 2)(x 4 + 11x 3 + 41x 2 + 61x + 30) T(x) = (x – 2)(x + 1)(x 3 + 10x 2 + 31x + 30) T(x) = (x – 2)(x + 1)(x + 2)(x 2 + 8x + 15) T(x) = (x – 2)(x + 1)(x + 2)(x + 3)(x + 5) x = {-5, -3, -2, -1, 2}

10. Homework 2.5 Zeros of Polynomial Functions page 160 1 - 31 odd, 37 - 85 odd, 91 - 94 all

11. #5 Find all of the zeros of the function. F(x) = (x + 6)(x + i)(x – i) x = {-6, -i, i}

12. #10 Use the rational zero test to list all of the possible rational zeros of f. Verify that the zeros of f shown are contained in the list. f(x) = 4x 5 – 8x 4 – 5x 3 + 10x 2 + x – 2

13. #15 Find all of the real zeros of the function. h(t) = t 3 + 12t 2 + 21t + 10 Caution graph may be misleading. With a cubic we expect another turn down to the left

14. #25 f(x) = x 3 + x 2 – 4x – 4 (a)List the possible rational zeros of f, (b)sketch the graph of f so that some possibilities can be eliminated (c)determine all the real zeros Graph eliminates -4, 1, 4 f(x) = x 3 + x 2 – 4x – 4 f(x) = x 2 (x +1)– 4(x + 1) f(x) = (x + 1)(x 2 – 4) f(x) = (x + 1)(x +2)(x – 2)

15. #42. Find the polynomial function with integer coefficients that has the given zeros. Since imaginary solutions always appear in conjugate pairs we know to include