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HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Hawkes Learning Systems: College Algebra Section 5.3: Locating Real Zeros of Polynomials HAWKES LEARNING SYSTEMS Copyright © 2011 Hawkes Learning Systems. All rights reserved. math courseware specialists Objectives o The Rational Zero Theorem. o Descartes’ Rule of Signs. o Bounds of real zeros. o The Intermediate Value Theorem. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. The Rational Zero Theorem o It can be proven (using ideas from a branch of mathematics called abstract algebra) that there is no formula based on elementary mathematical operations that identifies all the zeros of an arbitrary polynomial of degree five or higher. o However, there are some tools that give us hints about where to look for zeros of a given polynomial. One such tool is the Rational Zero Theorem. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. The Rational Zero Theorem If f x an x n an1x n1 ... a1x a0 is a polynomial with integer coefficients, then any p rational zero of f must be of the form , where q .p is a factor of the constant term a0 and q is a factor of the leading coefficient an . HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. The Rational Zero Theorem Caution The Rational Zero Theorem does not necessarily find even a single zero of a polynomial; instead it identifies a list of rational numbers that could potentially be zeros. In order to determine if any of the potential zeros actually are zeros, we have to test them. Additionally, the theorem says nothing about irrational zeros or complex zeros. If a given polynomial has some zeros that are either irrational or else have a non-zero imaginary part, we must resort to other means to find them. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 1: Factoring Polynomials For the following polynomial, list all of the potential rational zeros. Then write the polynomial in factored form and identify the actual zeros. f x 2 x3 9 x 2 17 x 6 Step 1: List the factors of the Factors of a0 : 1, 2,3,6 constant term, 6 , and the leading coefficient, 2 . Factors of a3: 1, 2 Step 2: By the Rational Zero Theorem, any rational zero must be one of the numbers generated by dividing factors of a0 by a3. Possible 1 3 Rational Zeros: 1, 2,3, 6, , 2 2 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 1: Factoring Polynomials (Cont.) Step 3: Use synthetic division to identify which of the potential zeros actually are zeros. You may have to try several potential zeros before finding an actual zero. Step 4: Once we determine that 1 is a zero, then we can finish by factoring 2 x 2 11x 6 . 1 2 9 17 6 2 11 6 2 11 6 0 f x 2 x 2 11x 6 x 1 2 x 1 x 6 x 1 1 Actual Zeros: ,6, 1 2 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 2: Factoring Polynomials For the following polynomial, list all of the potential rational zeros. Then write the polynomial in factored form and identify the actual zeros. g x 4 x 4 37 x3 2 x 2 93x 54 Factors of a0 : 1, 2,3,6,9,18, 27,54 Factors of a4 : 1, 2, 4 1 3 9 27 1 3 9 27 Possible Rational Zeros: 1, 2,3, 6,9,18, 27,54, , , , , , , , 2 2 2 2 4 4 4 4 HAWKES LEARNING SYSTEMS Copyright © 2011 Hawkes Learning Systems. All rights reserved. math courseware specialists Example 2: Factoring Polynomials g x 4 x 4 37 x3 2 x 2 93x 54 Use synthetic division to find the actual zeros. 3 4 4 37 2 93 54 3 30 21 54 2 4 40 28 72 8 64 72 0 4 32 36 0 3 g x 4 x 32 x 36 x 2 x 4 3 g x 4 x 9 x 1 x 2 x 4 3 Actual Zeros: 9,1, 2, 4 2 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Descartes’ Rule of Signs Let f x an x n an1x n1 ... a1x a0 be a polynomial with real coefficients, and assume a0 0. A variation in sign of f is a change in the sign of one coefficient of f to the next, either from positive to negative or vice versa. 1. The number of positive real zeros of f is either the number of variations in sign of f x or is less than this by a positive even integer. 2. The number of negative real zeros of f is either the number of variations in sign of f x or is less than this by a positive even integer. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Descartes’ Rule of Signs o Note that in order to apply Descartes’ Rule of Signs, it is critical to first write the terms of the polynomial in descending order. o Note also that unless the number of variations in sign is 0 or 1, the rule does not give us a definitive answer for the number of zeros to expect. o For example, if the number of variations in sign of .f x is 4 , we know only that there will be 4, 2, or 0 positive real zeros. HAWKES LEARNING SYSTEMS Copyright © 2011 Hawkes Learning Systems. All rights reserved. math courseware specialists Example 3: Using Descartes’ Rule of Signs Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of each of the following polynomials. Then use the Rational Zeros Theorem and other means to find the zeros, if possible. f x x3 6 x 2 3x 10 There are two variations in sign in the function f x , so we know there Sign Sign are 2 or 0 positive real zeros. change change f x x 6 x 3 x 10 3 2 x3 6 x 2 3x 10 Sign change There is only one variation in sign in the function .f x so we know there is exactly 1 negative real zero. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 3: Using Descartes’ Rule of Signs (Cont.) f x x3 6 x 2 3x 10 Factors of a0 : 1, 2,5,10 Factors of an : 1 Possible Rational Zeros: 1, 2,5,10 1 1 6 3 10 1 7 10 1 7 10 0 f x x 1 x 2 7 x 10 x 1 x 5 x 2 Zeros : 1,5, 2 Since we know there is exactly 1 negative real zero, use synthetic division to find this first. HAWKES LEARNING SYSTEMS Copyright © 2011 Hawkes Learning Systems. All rights reserved. math courseware specialists Example 4: Using Descartes’ Rule of Signs Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of the following polynomial. Then use the Rational Zeros Theorem and other means to find the zeros, if possible. There is exactly 1 f x 2 x3 6 x 2 10 x 30 positive real zero. Sign change f x 2 x 6 x 10 x 30 3 2 f x 2 x3 6 x 2 10 x 30 Sign change Sign change There are either 2 or 0 negative real zeros. HAWKES LEARNING SYSTEMS Copyright © 2011 Hawkes Learning Systems. All rights reserved. math courseware specialists Example 4: Using Descartes’ Rule of Signs (Cont.) f x 2 x3 6 x 2 10 x 30 Factors of a0 : 1, 2,3,5,6,10,15,30 Factors of an : 1, 2 1 3 5 15 Possible Rational Zeros: 1, 2,3,5, 6,10,15,30, , , , 2 2 2 2 3 2 6 10 30 0 30 6 0 2 0 10 f x x 3 2 x 2 10 2 x 3 x 5 x 5 Zeros: 3, 5, 5 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Upper and Lower Bounds of Zeros Assume f x is a polynomial with real coefficients, a positive leading coefficient, and degree 1 . Let a and b be fixed numbers, with a 0 b . Then: 1. No real zero of f is larger than b (we say b is an upper bound of the zeros of f ) if the last row in the synthetic division of .f x by x b contains no negative numbers. That is, b is an upper bound of the zeros if the quotient and remainder have no negative coefficients when f x is divided by x b. 2. No real zero of f is smaller than a (we say a is a lower bound of the zeros of f ) if the last row in the synthetic division of .f x by x a has entries that alternate in sign (0 can be counted as either positive or negative, as necessary). HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 5: Locating Bounds With Synthetic Division This example revisits the polynomial f x from Example 3. Use synthetic division to identify upper and lower bounds of the real zeros of the following polynomial. f x x3 6 x 2 3x 10 We can begin by trying out any positive number whatsoever as an upper bound. 5 1 6 3 10 5 5 10 Since there are negative numbers in the last row, 5 is not an upper bound. 1 1 2 0 6 1 6 3 10 6 0 18 1 0 3 28 The number 6 is an upper bound according to the theorem, as all of the coefficients in the last row are nonnegative. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 5: Locating Bounds With Synthetic Division 1 1 6 3 10 1 7 10 1 7 10 0 We find that 1 is a lower bound, as the signs in the last row alternate. Note: 0 can be positive or negative, as necessary. To summarize, we know that all real zeros of f lie in the interval 1,6, meaning there is no need to test the numbers 2, 5, 10,10 out of the list of potential rational zeros 1, 2,5,10 . We have already determined in Example 3 that the zeros of f are 1,5, 2 , all of which do indeed lie in the interval 1,6 . HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Upper and Lower Bounds of Zeros Caution Don’t read more into the Upper and Lower Bounds Theorem than is actually there, as the theorem is not always powerful enough to indicate the numbers that are the best upper or lower bounds. The tradeoff for this weakness in the theorem is that it is quickly and easily applied. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. The Intermediate Value Theorem o The last technique for locating zeros that we study makes use of a property of polynomials called continuity. o One consequence of continuity is that the graph of a continuous function has no “breaks” in it. o This means that, assuming that the function can be graphed at all, it can be drawn without lifting the drawing tool. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. The Intermediate Value Theorem Assume that f x is a polynomial with real coefficients, and that a and b are real numbers with a b . If f a and f b differ in sign, then there is at least one point c such that a c b and f c 0. That is, at least one zero of f lies between a and b . HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 6: Approximating Zeros Show that f x 5 x3 x 9 has a zero between 1 and 2. We could determine f 1 and f 2 1 5 0 1 9 by direct computation, but for 5 5 6 many polynomials, it is quicker to 5 5 6 3 use synthetic division as illustrated. 2 5 0 1 9 Remember that the remainder 10 20 42 upon dividing f x by x k is f k ; 5 10 21 33 thus f 1 3 and f 2 33. The critical point is that these two values are opposite in sign, so a zero of f must lie between 1 and 2. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 7: Approximating Zeros Find an approximation (to the nearest tenth) of the zero between 1 and 2 for the function from Example 6, f x 5x3 x 9 . Since the result of f 1.1 is 1.1 5 0 1 9 5.5 6.05 7.755 negative and the result of f 1.2 is positive, we know 5 5.5 7.05 1.245 that the actual zero must lie 1.2 5 0 1 9 between 1.1 and 1.2 . 6 7.2 9.84 The value of f midway 5 6 8.2 0.84 between 1.1 and 1.2 is 0 1 9 5.75 6.6125 8.754375 5 5.75 7.6125 0.245625 1.15 5 negative, so the zero must lie between 1.15 and 1.2 . We have shown that the zero, to the nearest tenth, is 1.2.