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Section 2.5 Part Two

Section 2.5 Part Two. Other Tests for Zeros Descartes’s Rule of Signs Upper and Lower Bounds. Descartes’s Rule of Signs. Let f(x) be a polynomial with real coefficients and a o ≠ 0.

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Section 2.5 Part Two

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  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 ao≠ 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) = x2 + 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 = x2 – 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) = x5 + 9x4 + 19x3 – 21x2 – 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) = -x5 + 9x4 – 19x3 – 21x2 + 92x – 60 T(-x) has four variation in signs

  5. Find the Zeros of T(x) • T(x) = x5 + 9x4 + 19x3 – 21x2 – 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) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60 T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30) 1 9 19 -21 -92 -60 +2 2 22 122 60 82 61 30 0 1 11 41 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) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60 T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30) • 30  -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30 T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30) 1 11 41 61 30 -1 -1 -10 -30 -31 30 0 1 10 31

  8. Find the Zeros of T(x) T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60 T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30) T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30) • 30  -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30 T(x) = (x – 2)(x + 1)(x + 2)(x2 + 8x + 15) 1 10 31 30 -2 -2 -16 -30 0 1 8 15

  9. Find the Zeros of T(x) T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60 T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30) T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30) T(x) = (x – 2)(x + 1)(x + 2)(x2 + 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) = 4x5 – 8x4 – 5x3 + 10x2 + x – 2

  13. #15 Find all of the real zeros of the function. h(t) = t3 + 12t2 + 21t + 10 Caution graph may be misleading. With a cubic we expect another turn down to the left

  14. #25 f(x) = x3 + x2 – 4x – 4 • List the possible rational zeros of f, • sketch the graph of f so that some possibilities can be eliminated • determine all the real zeros Graph eliminates -4, 1, 4 f(x) = x3 + x2 – 4x – 4 f(x) = x2 (x +1)– 4(x + 1) f(x) = (x + 1)(x2– 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

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