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The Rational Zero Theorem The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear somewhere in the list. The Rational Zero Theorem p If f(x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients and q p (where is reduced) is a rational zero, then p is a factor of the constant q term a0 and q is a factor of the leading coefficient an. EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f(x) = 15x3 + 14x2 - 3x – 2. Solution The constant term is –2 and the leading coefficient is 15. Factors of the constant term, - 2 Factors of the leading coefficient, 15 1, 2 = 1, 3, 5, 15 Possible rational zeros = = 1, 2, 13 , 23 , Divide 1 and 2 by 1. Divide 1 and 2 by 3. 15 , 52 , Divide 1 and 2 by 5. 1 , 2 15 15 Divide 1 and 2 by 15. There are 16 possible rational zeros. The actual solution set to f(x) = 15x3 + 14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions. • Verify that x = -1 is a root in the previous example… Integral Root Theorem • Suppose your leading coefficient in your polynomial function is 1. • What does this mean for you possible rational roots? • Ex: Find all the roots of… f ( x) = x + 8x + 16 x + 5 3 2 Descartes' Rule of Signs If f(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomial with real coefficients. 1. The number of positive real zeros of f is either equal to the number of sign changes of f(x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f is either equal to the number of sign changes of f(-x) or is less than that number by an even integer. If f(-x) has only one variation in sign, then f has exactly one negative real zero. EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f(x) = x3 + 2x2 + 5x + 4. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f(x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f(-x). We obtain this equation by replacing x with -x in the given function. f(x) = x3 + 2x2 + 5x + 4 Replace x with -x. f(-x) = (-x)3 + 2(-x)2 + 5(-x) + 4 = -x3 + 2x2 - 5x + 4 This is the given polynomial function. EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f(x) = x3 + 2x2 + 5x + 4. Solution Now count the sign changes. f(-x) = -x3 + 2x2 - 5x + 4 1 2 3 There are three variations in sign. # of negative real zeros of f is either equal to 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 - 2 = 1 negative real zero. Use Descartes’ Rule of signs to find the number of possible real roots… The, find all the roots. f ( x) = x - 2 x - 8 x 3 2