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3.3 Zeros of polynomial functions Rational Zero Test: If a polynomial f(x) has integer coefficients, every rational zero of f has the form p factors of the constant term rational zero = q factors of the L. coefficint where p and q have no common factors other than 1. • p is a factor of the constant term. • q is a factor of the leading coefficient. Example: List all possible rational zeros of f(x) = 4x3 + 3x2 – x – 6. q=4 p=–6 1 Find the zeros of f(x) = 2x3 + x2 – 5x + 2. f(x) = 2x3 – 9 x2 + 2. 2 4 2 2 3 f ( x) x 3x x 9 x 4 3 3 2 Upper and Lower Bounds Theorem • When P(x)÷(x-c) by syn. Div and P(x) with real coefficients. Upper Bound: a) c>0 and the leading coefficient > 0 x = c is an upper bound for the real zeros if all numbers in the bottom row are either +ve or zeros b) c>0 and the leading coefficient < 0 x = c is an upper bound for the real zeros if all numbers in the bottom row are either -ve or zeros Lower Bound: c<0 x = c is a lower bound for the real zeros if all numbers in the bottom row are alternate in sign ( the zero can be considered +ve or –ve) 3 Example: According to the Bounds Theorem find both the smallest positive integer that is upper bound and the largest negative integer that is lower bound for the real zeros • f (x) = – 4 x4 +12 x3+3x2 – 12x + 7. • f (x) = 4 x3 – 11x2 – 3x + 15. 4 Descartes’s Rule of Signs: If f(x) is a polynomial with real coefficients and a nonzero constant term, 1. The number of positive real zeros of f is equal to the number of variations in sign of f(x) or less than that number by an even integer. 2. The number of negative real zeros of f is equal to the number of variations in sign of f(–x) or less than that number by an even integer. A variation in sign means that two consecutive, nonzero coefficients have opposite signs. 5 Example: Use Descartes’s Rule of Signs to determine the possible number of positive and negative real zeros of f(x) = 2x4 – 17x3 + 35x2 + 9x – 45. The polynomial has three variations in sign. + to – + to – f(x) = 2x4 – 17x3 + 35x2 + 9x – 45 – to + f(x) has either three positive real zeros or one positive real zero. f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45 =2x4 + 17x3 + 35x2 – 9x – 45 f(x) has one negative real zero. One change in sign f(x) = 2x4 – 17x3 + 35x2 + 9x – 45 = (x + 1)(2x – 3)(x – 3)(x – 5). 6