Section 33 Approximating Real Zeros of Polynomials


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1  Approimating Real Zeros of Polynomials 9 Section  Approimating Real Zeros of Polynomials Locating Real Zeros The Bisection Method Approimating Multiple Zeros Application The methods for finding zeros discussed in the preceding section are designed to find as many eact real and imaginary zeros as possible. But there are zeros that cannot be found by using these methods. For eample, the polynomial P() must have at least one real zero (Theorem in Section ). Since the only possible rational zeros are, and neither of these turns out to be a zero, P() must have at least one irrational zero. We cannot find the eact value of this zero, but it can be approimated using various wellknown methods. In this section we develop two important tools for locating real zeros, the location theorem and the upper and lower bound theorem. We will see that the upper and lower bound theorem is a useful tool for approimating real zeros on a graphing utility. Net we discuss how the location theorem leads to the bisection method, a simple approimation technique that forms the basis for the zero approimation routines on most graphing utilities. Finally, we investigate some of the problems that can be encountered when a polynomial has multiple zeros. FIGURE P(). P() In this section, we restrict our attention to the real zeros of polynomials with real coefficients. Locating Real Zeros Let us return to the polynomial function P() As we have found, P() has no rational zeros and at least one irrational zero. The graph of P() is shown in Figure. Note that P() and P(). Since the graph of a polynomial function is continuous, the graph of P() must cross the ais at least once between and. This observation is the basis for Theorem and leads to a simple method for approimating zeros. THEOREM LOCATION THEOREM If f is continuous on an interval I, a and b are two numbers in I, and f(a) and f(b) are of opposite sign, then there is at least one intercept between a and b.
2 POLYNOMIAL AND RATIONAL FUNCTIONS We will find Theorem very useful when we are searching for real zeros, hence the name location theorem. It is important to remember that at least in Theorem means one or more. Notice in Figure (a) that f(), f(), and f has one zero between and. In Figure (b), f() and f(), but this time there are three zeros between and. FIGURE The location theorem. f() f() f() (a) f() (b) f() (c) f() The converse to the location theorem (Theorem ) is false; that is, if c is a zero of f, then f may or may not change sign at c. Compare Figures (a) and (c). Both functions have a zero at, but the first changes sign at and the second does not. EXAMPLE Solution FIGURE y P() 6 9. Locating Real Zeros Let P() 6 9. Use a table to locate the zeros of P() between successive integers and check graphically. We construct a table and look for sign changes [Fig. (a)]. The table in Figure (a) shows three sign changes. According to Theorem, P() must have a real zero in each of the intervals (, ), (, ), and (, ). The graph in Figure (b) confirms this. Since P() is a cubic polynomial, we have located all of its zeros. 6 (a) (b) MATCHED PROBLEM Let P() 8. Use a table to locate the zeros of P() between successive integers and check graphically. In the solution to Eample, we located three zeros in a relatively few number of steps and could stop searching because we knew that a cubic polynomial could not have more than three zeros. But what if we had not found three zeros?
3  Approimating Real Zeros of Polynomials Some cubic polynomials have only one real zero. How can we tell when we have searched far enough? The net theorem tells us how to find upper and lower bounds for the real zeros of a polynomial. Any number that is greater than or equal to the largest zero of a polynomial is called an upper bound of the zeros of the polynomial. Similarly, any number that is less than or equal to the smallest zero of the polynomial is called a lower bound of the zeros of the polynomial. Theorem, based on the synthetic division process, enables us to determine upper and lower bounds of the real zeros of a polynomial with real coefficients. THEOREM UPPER AND LOWER BOUNDS OF REAL ZEROS Given an nthdegree polynomial P() with real coefficients, n, a n, and P() divided by r using synthetic division:. Upper Bound. If r and all numbers in the quotient row of the synthetic division, including the remainder, are nonnegative, then r is an upper bound of the real zeros of P().. Lower Bound. If r and all numbers in the quotient row of the synthetic division, including the remainder, alternate in sign, then r is a lower bound of the real zeros of P(). [Note: In the lowerbound test, if appears in one or more places in the quotient row, including the remainder, the sign in front of it can be considered either positive or negative, but not both. For eample, the numbers,, can be considered to alternate in sign, while,, cannot.] We sketch a proof of part of Theorem. The proof of part is similar, only a little more difficult. Proof If all the numbers in the quotient row of the synthetic division are nonnegative after dividing P() by r, then P() ( r)q() R where the coefficients of Q() are nonnegative and R is nonnegative. If r, then r and Q() ; hence, P() ( r)q() R Thus, P() cannot be for any greater than r, and r is an upper bound for the real zeros of P(). Theorem requires performing synthetic division repeatedly until the desired patterns occur in the quotient row. This is a simple, but tedious, operation to carry out by hand. Table shows a simple program for a graphing calculator that will perform synthetic division.
4 POLYNOMIAL AND RATIONAL FUNCTIONS T A B L E Synthetic Division on a Graphing Utility Program SYNDIV TI8/TI8 TI8/TI86 Output Eplore/Discuss (A) If you have a TI8, TI8, TI8, or TI86 graphing calculator, enter the appropriate version of SYNDIV in your calculator eactly as shown in Table. If you have some other graphing utility that can store and eecute programs, consult your manual and modify the statements in SYNDIV so that the program works on your graphing utility. (B) Store the coefficients of P() 6 9 in L (see the first line of output in Table ) and eecute the program. Type at the R? prompt and press ENTER to display the results of synthetic division with r. To continue press ENTER again. Repeat this process for r,,, and. Press QUIT at the R? prompt to terminate the program. Compare the last number in each list with the values of P() in Figure (a). EXAMPLE Solution Bounding Real Zeros Let P() 9. Find the smallest positive integer and the largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). Also note the location of any zeros discovered in the process of searching for upper and lower bounds. We can use SYNDIV or hand calculations to perform synthetic division for r,,,... until the quotient row turns nonnegative; then repeat this process
5 afdbfc afdbfc FIGURE P() 9.  Approimating Real Zeros of Polynomials for r,,,... until the quotient row alternates in sign. We organize these results in the synthetic division table shown below. In a synthetic division table we dispense with writing the product of r with each coefficient in the quotient and simply list the results in the table. It is also useful to include r in the table to detect any sign changes between r and r. UB LB This quotient row is nonnegative; hence, is an upper bound (UB). This quotient row alternates in sign; hence, is a lower bound (LB). The graph of P() 9 for is shown in Figure. Theorem implies that all the real zeros of P() are between and. We can be certain that the graph does not change direction and cross the ais somewhere outside the viewing window in Figure. We also note that there is at least one zero in (, ) and at least one in (, ), as indicated by the sign changes in the values of P() shown in the last column of the synthetic division table. MATCHED PROBLEM Let P() 7. Find the smallest positive integer and the largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). Also note the location of any zeros discovered in the process of searching for upper and lower bounds. EXAMPLE Approimating Real Zeros with a Graphing Utility Let P() 7 7. (A) Find the smallest positive integer multiple of and the largest negative integer multiple of that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). (B) Approimate the real zeros of P() to two decimal places.
6 POLYNOMIAL AND RATIONAL FUNCTIONS Solutions (A) We construct a synthetic division table to search for bounds for the zeros of P(). The size of the coefficients in P() indicates that we can speed up this search by choosing larger increments between test values FIGURE P() 7 7. UB LB , 7,7 Thus, all real zeros of P() 7 7 must lie between and. (B) Graphing P() for (Fig. ) shows that P() has three zeros. The approimate values of these zeros (details omitted) are.8,.8, and.. MATCHED PROBLEM Let P() 7 7. (A) Find the smallest positive integer multiple of and the largest negative integer multiple of that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). (B) Approimate the real zeros of P() to two decimal places. Remark One of the most frequently asked questions concerning graphing utilities is how to determine the correct viewing window. The upper and lower bound theorem provides an answer to this question for polynomial functions. As Eample illustrates, the upper and lower bound theorem and the zero approimation routine on a graphing utility are two important mathematical tools that work very well together. The Bisection Method Now that we know more about locating real zeros of a polynomial, we want to discuss a technique that can be used to approimate real zeros. Eplore/Discuss provides an introduction to the repeated systematic application of the location theorem (Theorem ) called the bisection method. This method forms the basis for the zero approimation routines in many graphing utilities. Eplore/Discuss Let P(). Since P() and P(), the location theorem implies that P() must have at least one zero in (, ). (A) Is P(.) positive or negative? Is there a zero in (,.) or in (., )?
7  Approimating Real Zeros of Polynomials Eplore/Discuss continued (B) Let m be the midpoint of the interval from part A that contains a zero. Is P(m) positive or negative? What does this tell you about the location of the zero? (C) Eplain how this process could be used repeatedly to approimate a zero to any desired accuracy. The bisection method used to approimate real zeros is straightforward: Let P() be a polynomial with real coefficients. If P() has opposite signs at the endpoints of the interval (a, b), then a real zero r lies in this interval. We bisect this interval [find the midpoint m (a b)/], check the sign of P(m), and choose the interval (a, m) or (m, b) on which P() has opposite signs at the endpoints. We repeat this bisecting process (producing a set of nested intervals, each half the size of the preceding one and each containing the real zero r) until we get the desired decimal accuracy for the zero approimation. At any point in the process if P(m), we stop, since m is a real zero. An eample will help clarify the process. EXAMPLE Solution FIGURE 6 Nested intervals produced by the bisection method in Table. Approimating Real Zeros by Bisection For the polynomial P() 9 in Eample, we found that all the real zeros lie between and and that each of the intervals (, ) and (, ) contained at least one zero. Use bisection to approimate a real zero on the interval (, ) to onedecimalplace accuracy. We organize the results of our calculations in a table. Since the sign of P() changes at the endpoints of the interval (.6,.6), we conclude that a real zero lies in this interval and is given by r.6 to onedecimalplace accuracy (each endpoint rounds to.6)..6.6 ( (( ) ) )..7 Figure 6 illustrates the nested intervals produced by the bisection method in Table. Match each step in Table with an interval in Figure 6. Note how each interval that contains a zero gets smaller and smaller and is contained in the preceding interval that contained the zero. T A B L E Bisection Approimation Sign Change Interval (a, b) Midpoint m P(a) Sign of P P(m) P(b) (, ) (., ) (.,.7) (.,.6) (.6,.6) We stop here
8 6 POLYNOMIAL AND RATIONAL FUNCTIONS If we had wanted twodecimalplace accuracy, we would have continued the process in Table until the endpoints of a sign change interval rounded to the same twodecimalplace number. MATCHED PROBLEM Use the bisection method to approimate to onedecimalplace accuracy a zero on the interval (, ) for the polynomial in Eample. The bisection method is easy to implement on a graphing utility. Table shows a program that computes the sequence of nested intervals shown in Table. T A B L E Bisection on a Graphing Utility Program BISECT TI 8/8 TI 8/86 Output (a) (b) FIGURE 7 (c) Eplore/Discuss (A) If you have a TI8, TI8, TI8, or TI86 graphing calculator, enter the appropriate version of BISECT in your calculator eactly as shown in Table. If you have some other graphing utility that can store and eecute programs, consult your manual and modify the statements in BISECT so that the program works on your graphing utility.
9  Approimating Real Zeros of Polynomials 7 Eplore/Discuss continued (B) To use the program, first enter P() 9 as Y in the equation editor of the graphing utility [Fig. 7(a)]. Then enter and at the prompts and press ENTER repeatedly to generate the sequence of nested intervals [Fig. 7(b)]. Press ON and select Quit to terminate eecution [Fig. 7(c)]. FIGURE 8 P() ( ) ( ).. Approimating Multiple Zeros Consider the polynomial P() ( ) ( ) which has a simple zero at, a zero of multiplicity at and a zero of multiplicity at (see Fig. 8). Notice that P() has a local maimum at and does not change sign at. Also, has a local minimum at and does not change sign there either. Both of these are zeros of even multiplicity. On the other hand, is a zero of odd multiplicity, P() does change sign at, and does not have a local etremum at. Theorem, which we state without proof, generalizes these observations. THEOREM ZEROS OF EVEN AND ODD MULTIPLICITY If P() is a polynomial with real coefficients, then. If r is a zero of odd multiplicity, then P() changes sign at r and does not have a local etremum at r.. If r is a zero of even multiplicity, then P() does not change sign at r and has a local etremum at r. Eplore/Discuss Let P() ( ) ( ). (A) Use BISECT to approimate the zeros of P(). Where does it fail to work? Why? (B) Use the zero approimation routine on your graphing utility to approimate the zeros of P(). Does this routine ever fail to work? The bisection method shown in Table requires that the function change sign at a zero to approimate that zero. Thus, this method will always fail at a zero of even multiplicity. Most graphing utilities use a more complicated routine that may or may not work at zeros of even multiplicity. For eample, the TI8 graphing calculator was able to approimate the zero of P() ( ) ( ) at, but not the zero at. What are we to do if the zero routine fails? The ideas introduced in Theorem provide the answer:
10 8 POLYNOMIAL AND RATIONAL FUNCTIONS Zeros of even multiplicity can be approimated by using a maimum or minimum approimation routine, whichever applies. The net eample illustrates this approach. EXAMPLE Solution FIGURE 9 Zeros of P() 6 6. Approimating Multiple Zeros Let P() 6 6. Approimate the zeros of P() to two decimal places. Use maimum and minimum routines to approimate any zeros of even multiplicity. Determine the multiplicity of each zero. Eamining the graph of P(), we see that there is a zero of odd multiplicity at [Fig. 9(a)]. It appears that there may be zeros of even multiplicity near and. Using a maimum routine near [Fig. 9(b)] and a minimum routine near [Fig. 9(c)], we find that. and. are zeros of even multiplicity. Since P() is a fifthdegree polynomial, these zeros must be double zeros and must be a simple zero. (a) Simple zero (b) Double zero (c) Double zero MATCHED PROBLEM Let P() 6 7. Approimate the zeros of P() to two decimal places. Use maimum and minimum routines to approimate any zeros of even multiplicity. Determine the multiplicity of each zero. Application EXAMPLE 6 FIGURE Construction An oil tank is in the shape of a right circular cylinder with a hemisphere at each end (see Fig. ). The cylinder is inches long, and the volume of the tank is, cubic inches (approimately cubic feet). Let denote the common radius of the hemispheres and the cylinder. inches (A) Find a polynomial equation that must satisfy. (B) Approimate to one decimal place.
11  Approimating Real Zeros of Polynomials 9 Solutions (A) If is the common radius of the hemispheres and the cylinder in inches, then ( ) ( ) ( ) Volume of tank Volume of two hemispheres Volume of cylinder,, 6 6, Multiply by /. Thus, must be a positive zero of P() 6, (B) Since the coefficients of P() are large, we use larger increments in the synthetic division table: 6,,, UB,9 6, Graphing y P() for (Fig. ), we see that. inches (to one decimal place). FIGURE P() 6,. 7, 7, MATCHED PROBLEM 6 Repeat Eample 6 if the volume of the tank is, cubic inches. Answers to Matched Problems. Intervals containing zeros: (, ), (, ), (, 6). Lower bound: ; upper bound: 6; intervals containing zeros: (, ), (, ). (A) Lower bound: ; upper bound: (B) Real zeros:.,.6, (double zero), (simple zero),.6 (double zero) 6. (A) P() 6, (B).7 inches
12 POLYNOMIAL AND RATIONAL FUNCTIONS EXERCISE  A In Problems, use the table of values for the polynomial function P to discuss the possible locations of the intercepts of the graph of y P().... P() P() respectively, for the real zeros of P(). Also note the location of any zeros between successive integers. (B) If (k, k ) is the interval determined in part A that contains the largest real zero of P(), determine the number of additional intervals required by the bisection method to obtain a onedecimalplace approimation to this zero and state the approimate value of the zero.. P() 6. P() 7. P() 8. P() 9. P() P() 9 9. P(). P(). P() P() In Problems, (A) Find the smallest positive integer and largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). (B) Approimate the real zeros of each polynomial to two decimal places. In Problems 8, use a synthetic division table and Theorem to locate each real zero between successive integers.. P() 9 6. P() 9 7. P() 8. P() Find the smallest positive integer and largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of each of the polynomials given in Problems P(). P(). P() 9. P() 6 7. P(). P() B In Problems, (A) Find the smallest positive integer and largest negative integer that, by Theorem, are upper and lower bounds,. P() 8. P(). P() 7 6. P() 8 7. P() 8. P() 9. P(). P() 6 9 Problems refer to the polynomial P() ( ) ( )( ). Can the zero at be approimated by the bisection method? Eplain.. Can the zero at be approimated by the bisection method? Eplain.. Can the zero at be approimated by the bisection method? Eplain.. Which of the zeros can be approimated by a maimum approimation routine? By a minimum approimation routine? By the zero approimation routine on your graphing utility?
13  Approimating Real Zeros of Polynomials For each polynomial P() in Problems, use a maimum or minimum routine to approimate zeros of even multiplicity and a zero approimation routine to approimate zeros of odd multiplicity, all to two decimal places. State the multiplicity of each zero.. P() P() P() P() P() P() C In Problems, (A) Find the smallest positive integer multiple of and largest negative integer multiple of that, by Theorem, are upper and lower bounds, respectively, for the real zeros of each polynomial. (B) Approimate the real zeros of each polynomial to two decimal places.. P(). P() 7 7. P() 9,. P() 7,. P(),, 6. P() 7, 7. P(),7 7,87, 8. P() 9,8,67 8, 9. P().. 9,. P() ,6, APPLICATIONS Epress the solutions to Problems 6 as the roots of a polynomial equation of the form P() and approimate these solutions to three decimal places. Use a graphing utility, if available; otherwise, use the bisection method. inches on a side, will be cut from each corner, and then the ends and sides will be folded up (see the figure). Find the value of that would result in a bo with a volume of 6 cubic inches.. Manufacturing. A bo with a hinged lid is to be made out of a piece of cardboard that measures by inches. Si squares, inches on a side, will be cut from each corner and the middle, and then the ends and sides will be folded up to form the bo and its lid (see the figure). Find the value of that would result in a bo with a volume of cubic inches. in. 8 in. in. in.. Construction. A propane gas tank is in the shape of a right circular cylinder with a hemisphere at each end (see the figure). If the overall length of the tank is feet and the volume is cubic feet, find the common radius of the hemispheres and the cylinder. feet 6. Shipping. A shipping bo is reinforced with steel bands in all three directions (see the figure). A total of. feet of steel tape is to be used, with 6 inches of waste because of a inch overlap in each direction. If the bo has a square base and a volume of cubic feet, find its dimensions.. Geometry. Find all points on the graph of y that are unit away from the point (, ). [Hint: Use the distancebetweentwopoints formula from Section .]. Geometry. Find all points on the graph of y that are unit away from the point (, ).. Manufacturing. A bo is to be made out of a piece of cardboard that measures 8 by inches. Squares, y
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