The Real Zeros of a Polynomial Function

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Presentation transcript:

The Real Zeros of a Polynomial Function Section 5.2 The Real Zeros of a Polynomial Function

OBJECTIVE 1

(a) f(-2) = (-2)3 + 3(-2)2 + 2(-2) -1 = -1 (b) f(1) = (1)3 + 3(1)2 + 2(1) -1 = 5

a) f(-1) = -2(-1)3 - (-1)2 + 4(-1) + 3 = 0  x + 1 is a factor b) f(1) = -2(1)3 – (1)2 + 4(1) + 3 = 4  x - 1 is NOT a factor

OBJECTIVE 2

p: ±1, ±2, ±3, ±4, ±6, ±12 q: ±1, ±3 p/q: ±1, ±1/3, ±2, ±2/3, ±3, ±4, ±4/3, ±6, ±12

OBJECTIVE 3

p/q: ±1, ±1/3, ±2, ±2/3, ±3, ±4, ±4/3, ±6, ±12

f(-3) = 3(-3)3 + 8(-3)2 - 7(-3) – 12 = 0 f(-1) = 3(-1)3 + 8(-1)2 - 7(-1) – 12 = 0 f(4/3) = 3(4/3)3 + 8(4/3)2 - 7(4/3) – 12 = 0 -3 | 3 8 -7 -12 -9 3 12 ____________ 3 -1 -4 0 (x + 3)(3x2 –x - 4) (x + 3)(3x – 4)(x + 1)

p: ±1, ±3, ±9 q: ±1, ±2 p/q: ±1, ±1/2, ±3, ±3/2, ±9, ±9/2

p/q: -1, ±1/2, -3, -3/2, ±9, ±9/2

f(-3) = 2(-3)4 + 13(-3)3 + 29(-3)2 + 27(-3) + 9 = 0 -1 | 2 13 29 27 9 -2 -11 -18 -9 _________________ 2 11 18 9 0 (x + 1)(2x3 + 11x2 + 18x + 9) -3 | 2 11 18 9 -6 -15 -9 _________________ 2 5 3 0 (x + 1)(x + 3)(2x2 + 5x + 3) = (x + 1)(x + 3)(2x + 3)(x + 1) = (x + 1)2(x + 3)(2x + 3)

OBJECTIVE 5

BOUND

Smaller of two values is 10 Bound is -10 ≤ x ≤ 10 g(x) = 4(x5 – 1/2x3 + 1/2x2 + 1/4) Max{ 1, 3 + 9 + 5} = Max{1, 17} = 17 Max{ 1, 1/2 + 1/2 + 1/4} = Max{1, 5/4} = 5/4 1 + Max{3, 9, 5} = 1 + 9 = 10 1 + Max{1/2, 1/2, 1/4} = 1 + ½ = 3/2 Smaller of two values is 10 Bound is -10 ≤ x ≤ 10 Smaller of two values is 5/4. Bound is -5/4 ≤ x ≤ 5/4

May be at most 5 zeros. Can NOT use Rational Zeros Theorem (non-integer coefficients) Find bounds: Max{1, 1.8 + 17.79 + 31.672 + 37.95 + 8.7121 } = Max{1, 97.9241} = 97.9241 1 + Max{1.8, 17.79, 31.672, 37.95, 8.7121} = 38.95 Bound: 38.95

OBJECTIVE 6

Using the function from the last example, determine whether there is a repeated zero or two distinct zeros near 3.30. f(3.29) = .00956 > 0 and f(3.30) = -.0001 < 0 f(3.30) = -.0001 < 0 and f(3.31) = .0097 > 0

Use xmin 3.29 and xmax 3.31 Use ymin -.001 and ymax + .001