2.5 Real Zeros of Polynomial Functions Descartes Rule of Signs

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Presentation transcript:

2.5 Real Zeros of Polynomial Functions Descartes Rule of Signs The number of positive real zeros of f is either equal to the number of sign changes of f(x) or is less by an even integer. The number of negative real zeros of f is either equal to the number of sign changes of f(-x) or is less by an even integer.

Ex. Apply Descartes’s Rule of Signs to f(x) = 2x4 + 7x3 – 4x2 – 27x - 18 How many sign variations are there in f(x)? Only one sign change, so f must have exactly one positive real zero. How about f(-x) f(-x) = 2x4 – 7x3 – 4x2 + 27x -18 3 variations in sign means f has either 3 or 1 negative real zeros.

Apply Descartes’s Rule to f(x) = x3 – x + 1 2 or 0 (+) - + + 1 (-) The Rational Zero Test Rational Zero = p = a factor of the constant term q = a factor of the leading coefficient

Find the rational zeros of Descartes Rule 3 or 1 (+) Ex. f(x) = x4 – x3 + x2 – 3x - 6 1 (-) factors of 6 Possible rational zeros = factors of 1 Start with 1 1 -1 1 -3 -6 1 1 1 -2 -8 did not = 0, so not a zero 2 1 1 3 3 Works! (x – 2) is a factor -1 1 3 Works! (x + 1) is a factor (x2 + 3)(x – 2)(x + 1) are the factors & -1 and 2 are zeros

Ex. Find the rational zeros of f(x) = 2x3 + 3x2 – 8x + 3 Desc. 2 or 0 (+) Poss. zeros 1 (-) 2 3 -8 3 1 2 5 -3 It works! (x – 1)(2x2 + 5x –3) (x – 1)(2x – 1)(x + 3) 1 ½ -3

Ex. f(x) = 10x3 – 15x2 – 16x + 12 Desc. 2 or 0 (+) Poss. Zeros 1 (-) 10 -15 -16 12 1 10 -5 -21 -9 2 10 5 -6 Works! (x – 2)(10x2 + 5x – 6) Quad. form

Use the given zero to find all the zeros of the function. Given: 3i f(x) = x3 + x2 + 9x + 9 If 3i is a zero, then -3i, it’s conjugate, is a zero. Multiply these zeros in factored form. (x - 3i)(x + 3i) = x2 + 9 Zeros are 3i, -3i, and -1