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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 ao ≠ 0. f ( x ) a n x a n 1 x n n 1 ... a 2 x a1 x a 0 2 • 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. 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} 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 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 Find the Zeros of T(x) T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60 T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30) +2 1 1 9 19 -21 -92 -60 2 22 82 122 11 41 61 30 60 0 Now that we have found the positive root we know that any other real roots must be negative. 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 1 1 11 41 61 -1 -10 -31 -30 10 31 30 30 0 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) -2 1 1 10 31 30 -2 -16 -30 8 15 0 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} Homework 2.5 • • • • • Zeros of Polynomial Functions page 160 1 - 31 odd, 37 - 85 odd, 91 - 94 all #5 Find all of the zeros of the function. F(x) = (x + 6)(x + i)(x – i) x = {-6, -i, i} #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 4 Factors of 2 Factors of 4 2 1 2 1 2 4 1 2 -2 -4 -6 1 2 1 4 1 1 x { 1, , , 1, 2} 2 2 #15 Find all of the real zeros of the function. h(t) = t3 + 12t2 + 21t + 10 Factors of 10 Factors of 1 1 2 5 10 1 Caution graph may be misleading. 6 With a cubic we expect another turn down to the left 4 2 x { 10 , 1, 1} -5 5 -2 #25 f(x) = x3 Factors of 4 Factors of 1 + x2 – 4x – 4 1 2 4 1 (a) (b) (c) List the possible rational zeros of f, sketch the graph of f so that some possibilities can be eliminated determine all the real zeros 1, 1, 2 , 2 , 4 , 4 4 Graph eliminates -4, 1, 4 f(x) = x3 + x2 – 4x – 4 2 f(x) = x2 (x +1)– 4(x + 1) -5 5 f(x) = (x + 1)(x2– 4) -2 f(x) = (x + 1)(x +2)(x – 2) -4 -6 x { 2 , 1, 2} #42. Find the polynomial function with integer coefficients that has the given zeros. 5, 5, 1 Since imaginary solutions always appear in conjugate pairs we know to include 3i 1 f ( x ) ( x 5 )( x 5 )( x 1 3 i )( x 1 f ( x ) ( x 25 )( x 2 x 4 ) 2 2 f ( x ) x 2 x 21 x 50 x 100 4 3 2 3i 3i )