* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Section 3.2
Survey
Document related concepts
Big O notation wikipedia , lookup
List of important publications in mathematics wikipedia , lookup
Proofs of Fermat's little theorem wikipedia , lookup
Fundamental theorem of calculus wikipedia , lookup
Horner's method wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Elementary mathematics wikipedia , lookup
Vincent's theorem wikipedia , lookup
Factorization of polynomials over finite fields wikipedia , lookup
Transcript
pg158 [V] 158 G2 5-36058 / HCG / Cannon & Elich Chapter 3 clb 11-22-95 MP1 Polynomial and Rational Functions 3.2 LOCATING ZEROS The man who breaks out into a new era of thought is usually himself still a prisoner of the old. Even Isaac Newton, who invented the calculus as a mathematical vehicle for his epoch-making discoveries in physics and astronomy, preferred to express himself in archaic geometrical terms. Freeman Dyson I had a good teacher for freshman algebra. I think he was simultaneously the football coach. Then I took sophomore geometry. It was apparently thought that students couldn’t learn geometry in one year so they had a second course in the junior year. The teacher in this second course didn’t understand the subject and I did. I made a lot of trouble for her. Saunders MacLane In Section 3.1 we indicated that two of the important concerns for polynomials are locating zeros and local extrema. With calculator graphs we can make excellent approximations for both. On a graph, there is no obvious relation between zeros and turning points, but in calculus we learn that every turning point of a polynomial function f occurs at a zero of another polynomial function called the derivative of f. Thus the location of turning points also depends on locating zeros. We leave the study of derivatives to calculus, but we devote this section to understanding more about zeros of polynomial functions. It would be nice to have something analogous to the quadratic formula for higher degree polynomials. For some polynomial functions, zeros can be expressed in exact form using radicals and the ordinary operations of algebra, but in general, we must rely on approximations. See the two Historical Notes in this section. Some of the theorems included in this section help us determine whether or not exact form solutions are available. Locator Theorem Graphs of all polynomial functions share some common properties; they are continuous and smooth, with no corners, breaks, or jumps. The idea of continuity (no breaks or jumps) is another topic for calculus and subsequent courses. We need some such theorem because calculator graphs are neither smooth nor continuous. Every calculator graph is the result of computing lots of discrete function values (one for each column of pixels). Dot mode shows only the isolated points. How are we to be confident that there isn’t some break or jump in the graph between two adjacent pixel columns? In connected mode, a graphing calculator connects different points of the graph by vertical strips, which we know cannot be part of the graph of any function. Nevertheless, our eye smooths out calculator graphs. We have come to expect graphs to be smooth and continuous, and the following theorem supports our intuition. It says, in effect, that a polynomial function cannot change from positive to negative without going through 0. The locator theorem is a special case of a theorem from analysis called the Intermediate Value Theorem. Locator (sign-change) theorem Suppose p is a polynomial function and a and b are numbers such that p~a! and p~b! have opposite signs. The function p has at least one zero between a and b, or equivalently, the graph of y 5 p~x! crosses the x-axis between ~a, 0! and ~b, 0!. cEXAMPLE 1 Using the locator theorem Let p~x! 5 2x 3 2 2x 2 2 3x 1 1. (a) Use the locator theorem to verify that p has a zero between 0 and 1. (b) Using a decimal window, graph y 5 p~x! and trace to verify that the zero is located between 0.2 and 0.3. pg159 [R] G1 5-36058 / HCG / Cannon & Elich jb 3.2 11-13-95 QC2 Locating Zeros 159 Solution (.2, .336) (a) p~0! 5 1 and p~1! 5 22, so there is a sign change and hence a zero between 0 and 1. (b) Tracing along the graph of p~x! in a decimal window, we can read the coordinates shown in Figure 11. Thus p~0.2! 5 0.336 and p~0.3! 5 20.026. There is a sign change between 0.2 and 0.3, so there is a zero in the interval. b (.3, .026) [– 5, 5] by [– 3.5, 3.5] p(x) = 2x3 – 2x2 – 3x + 1 FIGURE 11 We could, of course, use the calculator to locate the zero in Example 1 more precisely. If, for example, we simply zoom in on the point (0.2, 0) and then trace, we can locate the zero between 0.275 and 0.30. With time and patience we can locate zeros with as much accuracy as a calculator will display. A closer approximation is 0.2929. To get zeros in exact form, to move beyond approximations, we need other tools. The most powerful technique available to us involves factoring and the zeroproduct principle. Unfortunately, in most cases where we need the zeros of a given polynomial function, there is no dependable procedure for finding even one zero in exact form, and with polynomial functions of higher degree, even knowing several zeros may not be enough. The Division Algorithm Just as we can divide one integer by another to get an integer part q and a remainder r, where the remainder must be smaller than the divisor, so we can divide one polynomial by another. The result of polynomial division is a polynomial part q~x!, and a remainder r~x! whose degree must be smaller than the degree of the divisor. In particular, when the divisor is a linear polynomial (of the form x 2 c), then the remainder is some number r. This result is stated as a theorem known as the Division Algorithm. Properly, in the statement of the theorem, the degree of the divisor, d~x!, must be no greater than the degree of the polynomial we are dividing, p~x!. In our work, we assume a divisor that is either a linear or a quadratic polynomial. Division algorithm If p~x! is a polynomial of degree greater than zero, and d~x! is a polynomial, then dividing p~x! by d~x! yields unique polynomials q~x! and r~x!, called the polynomial part and remainder, respectively, such that p~x! 5 d~x! · q~x! 1 r~x!, where the degree of r~x! is smaller than the degree of d~x!. If d~x! 5 x 2 c, then the remainder is a unique number r, such that p~x! 5 ~x 2 c! · q~x! 1 r. (1) To find the polynomial part and remainder for any given pair of polynomials we use the familiar process of long division. For the special case of a linear divisor, there is a shortcut called synthetic division. Synthetic division is stressed in traditional courses because it is also used for several different evaluation purposes. With graphing calculators, however, the convenience of synthetic division does not justify the time required to learn the process. We outline synthetic division in the following (optional) discussion. You may divide by any method you wish, but we will do all of our polynomial division by long division. pg160 [V] 160 G6 5-36058 / HCG / Cannon & Elich Chapter 3 clb 11-22-95 MP1 Polynomial and Rational Functions HISTORICAL NOTE IS THERE A CUBIC FORMULA? The Babylonians could solve some quadratics nearly four thousand years ago, as could the ancient Greeks and Egyptians although they thought only positive roots had meaning. In essence, the quadratic formula has been around for at least a thousand years. From at least 1200 A.D. people have searched for a comparable formula for cubics. The story of Although Cardan is known for who first succeeded, and when, gets developing the first cubic formula, credit actually muddled by conflicting claims. At belongs to Tartaglia, pictured least part of the credit belongs to here. Scipione del Ferro (ca. 1510). By about 1540, Tartaglia had learned enough to win a public contest, solving 30 cubics in 30 days. x5 Somehow, Cardan got the method from Tartaglia and published it in 1545, much to Tartaglia’s dismay. The solution is often called Cardan’s even though he did credit Tartaglia. The methods of this chapter are much easier to apply, but the formula from Cardan’s book still works. Given a cubic of the form x 3 1 ax 1 b 5 0, first calculate A5 SD SD a 3 3 1 b 2 . 2 Cardan’s solution is given by Î 3 ÏA 2 b 2 2 Î 3 ÏA 1 b . 2 Synthetic Division Algorithm (Optional) Synthetic division is sometimes a convenient method for factoring a polynomial function. We do not justify the steps, but the procedure is really nothing but a short-cut method of dividing, using only the coefficients. The steps are outlined in the following box. Synthetic division algorithm (for divisors of the form x 2 c) To divide a polynomial p~x! of degree n by x 2 c: 1. On the top line write c (change sign from x 2 c), followed by all the coefficients of p~x! in order of decreasing powers of x, including any zero coefficients. 2. Bring down the leading coefficient, multiply by c, and add the product to the next coefficient to get the next entry on the bottom line. Repeat, multiplying the sum by c, and adding the product to the next coefficient, and continue for all coefficients of p. 3. The first n numbers on the bottom line are the coefficients of q~x!, of degree n 2 1, and the last number is the remainder r. We illustrate the synthetic division algorithm with the problem from Example 2: Divide p~x! 5 2x 3 2 2x 2 2 3x 1 1 by x 1 1. You may want to compare the coefficients in the long division of Example 2 with the numbers in the synthetic division. Since x 1 1 5 x 2 ~21!, We write 21 on the top line at the left, pg161 [R] G1 5-36058 / HCG / Cannon & Elich clb 11-22-95 MP1 3.2 Locating Zeros 161 followed by the coefficients of p~x! in order of decreasing powers of x: c from x 2 c A 21 22 23 1 22 4 21 2 24 1 0 2 q~x! 5 2x 2 2 4x 1 1 r50 B Coefficients of p~x! B For each entry on middle line, multiply bottom entry by 21, and add. From the last line we read the coefficients of the polynomial part q~x!, when p~x! is divided by x 1 1, and the remainder r. Writing p~x! in the form of the division algorithm, 2x 3 2 2x 2 2 3x 1 1 5 ~x 1 1!~2x 2 2 4x 1 1! 1 0. cEXAMPLE 2 Using the division algorithm Verify that 21 is a zero of the polynomial function from Example 1, p~x! 5 2x 3 2 2x 2 2 3x 1 1, and find the other two zeros in exact form. Solution Substituting 21 for x, p~21! 5 0. Dividing p~x! by x 1 1 yields the following: 2x 2 2 4x 1 1 x 1 1 ) 2x 3 2 2x 2 2 3x 1 1 2x 3 1 2x 2 24x 2 2 3x 24x 2 2 4x x11 x11 0 Since the remainder is 0, p~x! can be written in factored form as 2x 3 2 2x 2 2 3x 1 1 5 ~x 1 1!~2x 2 2 4x 1 1!. By the zero-product principle, either x 1 1 5 0 or 2x 2 2 4x 1 1 5 0. Using the quadratic formula for the second equation, we find that the zeros of p are 21, 2 2 Ï2 2 1 Ï2 , 2 . The second zero is about 0.29289, clearly the number we were 2 “zeroing in on” in Example 1. b cEXAMPLE 3 The division algorithm again (a) Show that x 2 1 4 is a factor of the polynomial (b) Find all zeros of p~x!. p~x! 5 x 4 2 x 3 1 2x 2 2 4x 2 8. Solution (a) We use long division. x2 2 x 2 2 x 1 4 ) x 4 2 x 3 1 2x 2 2 4x 2 8 1 4x 2 x4 3 2 x 2 2x 2 2 4x 2 4x 2 x3 28 2 2x 2 28 2 2x 2 0 2 pg162 [V] 162 G2 5-36058 / HCG / Cannon & Elich Chapter 3 clb 11-16-95 MP1 Polynomial and Rational Functions By the division algorithm, since the remainder is 0, p~x! 5 ~x 2 1 4!~x 2 2 x 2 2!, and we have a factorization of p~x!, thus reducing the problem of finding the zeros of a fourth polynomial function p to solving two quadratic equations, x 2 1 4 5 0, and x 2 2 x 2 2 5 0. (b) The four zeros of p~x! are 6 2i, 21, and 2. With two real zeros, the graph of y 5 p~x! should cross the x-axis just twice. See Figure 12. b Factor and Remainder Theorems [– 2, 3] by [– 10, 2] p(x) = x 4 – x 3 + 2x 2 – 4x – 8 FIGURE 12 Examples 2 and 3 illustrate one of the key concepts in finding exact form zeros of polynomial functions. Finding a zero is equivalent to finding a factor, and once we have factors, by the zero-product principle, the zeros of p are the zeros of the factors. Further, since the degree of q is smaller than the degree of p, q~x! is called the reduced polynomial. In the case where the divisor is linear, the division algorithm provides two powerful theorems. Consider again Equation (1), p~x! 5 ~x 2 c!q~x! 1 r. First, suppose r 5 0. Then ~x 2 c! is a factor, p~x! 5 ~x 2 c!q~x!. Conversely, if ~x 2 c! is a factor of p~x!, then p~x! 5 ~x 2 c!q~x!, so r must be 0. Equation (1) is an identity, so we can replace x by any number and obtain a true statement. In particular, if we replace x by c, we obtain p~c! 5 ~c 2 c!q~c! 1 r 5 0 · q~c! 1 r 5 0 1 r 5 r. Thus the remainder always equals the value of the function p at the number c. Putting these observations together, we have the following. Remainder and factor theorems When p~x! is divided by x 2 c, the remainder is p~c!. When p~x! is divided by x 2 c, then x 2 c is a factor of p~x! if and only if p~c! 5 0. The factor and remainder theorems give us ways to find a remainder without performing a lengthy division and can simplify many calculations. cEXAMPLE 4 Strategy: By the remainder theorem, for (a) p~21! 5 r, and for (b) P~22! 5 0, from which we can solve for k. The remainder theorem (a) Find the remainder when the polynomial p~x! 5 4x 15 1 5x 7 1 2x 4 1 3 is divided by x 1 1. (b) Find the value of k such that if the polynomial P~x! 5 x 3 1 x 2 1 kx 2 4 is divided by x 1 2, then the remainder is 0. Solution (a) Following the strategy, p~21! 5 4~21! 1 5~21! 1 2~1! 1 3 5 24. Hence, p~21! 5 24, so r 5 24. (b) Evaluating P at 22, we have P~22! 5 28 1 4 2 2k 2 4 5 2822k. Since the remainder when P~x! is divided by x 1 2 is 0, then P~22! 5 0. Thus, 28 2 2k 5 0, or 22k 5 8, or k 5 24. The desired function is P~x! 5 x 3 1 x 2 2 4x 2 4. b pg163 [R] G5 5-36058 / HCG / Cannon & Elich HISTORICAL NOTE clb 11-22-95 MP1 3.2 Locating Zeros 163 THERE IS NO “QUINTIC FORMULA” Very shortly after discovery of the general solution of the cubic equation (see “Is There a Cubic Formula?”), Ferrari (Italy, ca. 1545) derived a method for quartics (polynomials of degree 4). For n 5 2, 3, or 4, solutions for equations of degree n involve nth roots. Why not for degree 5? Nearly three hundred years passed before much more was done. Then, within a few years, two brilliant young men completely resolved the question. In 1820 Niels Henrik Abel of Norway was 18 when he thought he had the desired formula. Before it could be checked by others, however, he found his error and proved that there could be no general solution for quintic equations. In Paris in 1829 another 18-year-old, Evariste Galois, took the final step. In papers written during 1829 and 1830, Galois found the conditions that determine just which polynomial equations of degree 5 or higher can be solved in terms of their coefficients. Abel died of tuberculosis in 1829 at age 26. In 1831, at the age of 20, Galois was killed in a duel he himself recognized as stupid. During their brief careers, they laid the foundations for modern group theory, which has applications as diverse as solutions for Rubik’s cube and the standard model of elementary particle physics at the beginning of the universe. Clearing Fractions and Rational Zeros Multiplying an equation by a nonzero constant to clear fractions does not change the roots of the equation. For example, multiplying both sides of the equation 2 7 7 x 3 1 x 2 1 x 2 5 0, 3 2 3 by 6, we get an equation with integer coefficients having the same roots, 6x 3 1 21x 2 1 14x 2 4 5 0. If all coefficients are integers, then the following theorem provides a complete list of all the rational numbers that can possibly be zeros. Rational zeros theorem Let p be any polynomial function with integer coefficients. The only rational numbers that can possibly be zeros of p are the numbers of the form rs , where r is a divisor of the constant term, and s is a divisor of the leading coefficient. If none of these numbers is a zero, then p has no rational zeros. The rational zeros theorem is useful and important because it lists all the possibilities for rational zeros. The theorem does not tell us whether a given polynomial has any rational zeros at all; many do not. Without graphing technology, the theorem is a great help in guiding the search for zeros. Used with a grapher, the theorem can tell us things about graphs that the calculator cannot. No calculator can distinguish between rational and irrational numbers; every decimal is pg164 [V] 164 G2 5-36058 / HCG / Cannon & Elich Chapter 3 jb 11-13-95 QC2 Polynomial and Rational Functions truncated (“cut off ”) to fit the display capacity. Knowing what the rational possibilities are, we can use a calculator to verify that a particular zero is or is not a rational number. cEXAMPLE 5 Rational possibilities Use the rational zeros theorem to list all possible zeros of the polynomial function. (a) P~x! 5 x 3 2 4x 2 1 x 2 6 Strategy: (b) To get integer coefficients, multiply through by 9 and then apply the rational zeros theorem. Zeros of 9R are are the same as the zeros of R. (b) R~x! 5 x 4 2 4x 3 1 14 9 x2 1 44 9 x2 5 3 Solution (a) For P, begin by listing all possible numerators (factors of 26) and denominators (factors of 1): Possible numerators: Possible denominators: 6 6 1, 2, 3, 6 1 The only possibilities for rational zeros of P are the integers 26, 23, 22, 21, 1, 2, 3, and 6. (b) Follow the strategy and find all possible rational zeros of S~x! 5 9R~x!: S~x! 5 9x 4 2 36x 3 1 14x 2 1 44x 2 15. Possible numerators are factors of 15; denominators are factors of 9. Possible numerators: Possible denominators: 6 1, 3, 5, 15 6 1, 3, 9 The rational zeros theorem tells us that S, and hence R, has only sixteen possible rational zeros (in reduced form): F 6 1, 3, 5, 15, 13 , 53 , 19 , 5 9 G . b Having listed lots of possibilities for rational zeros of the polynomials P and R in Example 5, what do we know of the actual zeros? At this stage, we have nothing but possibilities. When we add graphing technology, we can say a great deal more, as in the next example. cEXAMPLE 6 Finding rational zeros Find all rational zeros, if there are any, of the polynomial functions P and R in Example 5. Approximate any irrational zeros to two decimal place accuracy. (a) P~x! 5 x 3 2 4x 2 1 x 2 6 (b) R~x! 5 x 4 2 4x 3 1 14 9 x2 1 44 9 x2 5 3 Solution (a) We don’t know anything about P except that as a cubic it must have at least one real zero. We begin with a graph in the @210, 10# 3 @210, 10# window. See Figure 13a. It is clear that there is exactly one real zero, very near 4. Looking at the list of rational zeros for P from Example 5, we see that the only positive rational possibilities are 1, 3, and 6. Therefore, the real zero of the polynomial P cannot be a rational number. If we zoom into a very small box around the pg165 [R] G1 5-36058 / HCG / Cannon & Elich clb 11-14-95 QC4 3.2 Locating Zeros 165 8 4 –8 –4 4 8 4 4.05 4.10 4.15 –4 –8 [3.8, 4.2] by [– 0.1, 0.1] (b) [– 10, 10] by [– 10, 10] (a) FIGURE 13 P~x! 5 x 3 2 4x 2 1 x 2 6 x-intercept point of the graph (Figure 13b) and then trace, we find that the zero is very near 4.11. (b) For graphing purposes, we can choose either the polynomial R~x!, or the polynomial S~x! 5 9R~x! with integer coefficients, because R and S have the same zeros. If we were working by hand, most of us would choose S to avoid fractions; for the calculator there is no difference, except that R requires a smaller window. The graph of R is shown in the @210, 10# 3 @210, 10# window we used for part (a) in Figure 14a. When we zoom into a box just large enough to include the zeros of the graph, as in Figure 14b, we can trace along the curve to find zeros near 21, 0.3, 1.6, and 3. –1 [– 10, 10] by [– 10, 10] (a) 1 2 3 [– 1.5, 4] by [– 3.2, 3] (b) FIGURE 14 Which, if any, of these are rational zeros? The list of rational possibilities (from Example 5) includes 21, 3, 13 , and 53 , all of which are reasonable candidates, but are these actually zeros of the function? To decide, we need to evaluate R at each number (see the following Technology Tip). To calculator accuracy we find that R~21! 5 0, R~3! 5 0, R~13 ! 5 0, R~53 ! 5 0. We conclude that R has four rational zeros, 21, 3, 13 , and 53 . b pg166 [V] 166 G2 5-36058 / HCG / Cannon & Elich Chapter 3 jb 11-13-95 QC2 Polynomial and Rational Functions TECHNOLOGY TIP r Function evaluation Most calculators, when we trace along a curve, display coordinates. The y-coordinate is the calculated value corresponding to the x-coordinate of the pixel, but we have no way to specify a particular x-value unless our window happens to have it as a pixel. If our goal is to evaluate R~ 13 !, we don’t want to settle for R (.324076113). We give here some suggestions for different calculators, but there may be a more efficient way for your particular machine. The displayed value is the calculator’s evaluation of R~ 13 !, which may involve round-off error. For example, one of our calculators displays R~ 13 !, 5 3E-13, meaning 0.0000000000003, and which in this context we interpret as 0. TI calculators: If you are graphing a function as Y1, return to the home screen, store the desired value in the x-register, and then enter Y1 . Thus, for R~21!, 21A X Enter. Then call up Y1 from the Y-vars menu (or on the TI-85, 2nd Alpha y1, and Enter. The TI-82 will evaluate Y1(21) directly. Casio calculators: The function must be entered on your MEM list, so type in the function, SHIFT MEM, F1(STO) and the number, say 1 for f1 .To evaluate f1~3!, EXIT and store 3 in the x-register, 3 A X EXE . Then F2(RCL) 1 EXE. HP–38: Having entered a function as, say, F1~X!, return to the home screen, type F1~1y3!, and ENTER. HP-48: The calculator will evaluate the function at any pixel-address and store the result on the stack, but direct evaluation is less convenient. One way is to store the number in, say, register A: 21 ENT ‘A’ STO. Then write the function as an expression in A: ‘A ` 4 2 4*A ` 3 1 14/9*A ` 2 1 44/9*A 2 5/3’. ENTER twice, so you have an extra copy on the stack. Then, purple NUM converts to a number. For another value, store it in A, Enter, and evaluate as a number. All these calculators except the TI-81 have some sort of SOLVE routine that will approximate zeros directly, which does not make it less important for you to understand the ideas we are discussing here. Applying the Rational Zeros Theorem The rational zeros theorem may be applied to problems other than looking for the zeros of a particular polynomial function. For instance, if we know that some number c is a zero of a polynomial function but c is not among the possibilities for rational zeros, then we can conclude that c is not a rational number, as in the next example. cEXAMPLE 7 Showing a number is not rational (a) Find a polynomial equation with integer coefficients satisfied by the number 3 c 5 Ï5 2 1. (b) Use the rational zeros theorem to show that c is not a rational number. Solution (a) The simplest polynomial equation satisfied by c is 3 x 5 Ï5 2 1, pg167 [R] G1 5-36058 / HCG / Cannon & Elich jb 3.2 11-13-95 QC2 Locating Zeros 167 but of course, this does not have integer coefficients. To get integer coefficients, we add 1, cube both sides, and simplify: @See the inside front cover for the expansion of ~x 1 1!3.] 3 ~x 1 1!3 5 ~Ï5!3 x 3 1 3x 2 1 3x 1 1 5 5 x 3 1 3x 2 1 3x 2 4 5 0. Now we have a polynomial equation, one of whose solutions is the number c. Incidentally, a graph makes it clear that p~x! 5 x 3 1 3x 2 1 3x 2 4 has only 3 one real zero, which thus must be Ï5 2 1. (b) Since the leading coefficient of p~x! is 1, the only possible rational zeros of p are 6 @1, 2, 4#. The number c is a zero of p but is not one of the possible 3 rational zeros, so Ï5 2 1 must be an irrational number. b (.5, 0) (– 1, 0) (2, 0) [– 5, 5] by [– 3.5, 3.5] 2x 3 – 3x 2 – 3x + 2 $ 0 on [– 1, 0.5] < [2, `) cEXAMPLE 8 Solving an inequality 2x 3 2 3x 2 2 3x 1 2 $ 0. Find the solution set for Solution A graph of f ~x! 5 2x 3 2 3x 2 2 3x 1 2 is shown in Figure 15. It appears from the graph that the zeros are 21, 12 , and 2, as we may verify, either by tracing in a decimal window, or by evaluating the function. The function f is clearly positive between 21 and 12 , and whenever x . 2. Therefore, the solution set for the inequality is given by FIGURE 15 S 5 @21, 0.5# < @2, `!. b EXERCISES 3.2 Check Your Understanding Draw a graph whenever helpful. Exercises 1–6 True or False. Give reasons. 1. The function p~x! 5 4x 3 2 x has three real zeros. 2. The positive zero of f ~x! 5 x 3 2 3x is less than 1.73. 3. For f ~x! 5 x 3 2 1.6x 2 2 8.52x 1 15.84, since f ~2! and f ~3! are positive, then f contains no zeros between 2 and 3. 4. The equation 2x 3 2 5x 2 1 4x 2 1 5 0 has no rational roots. 5. The function f ~x! 5 ~3x 2 2!~x 2 2 2x 2 4! has exactly one real zero. 6. When x 3 2 2x 2 1 3x 2 16 is divided by x 2 3, then the remainder is 2. Exercises 7–10 Fill in the blank so that the resulting statement is true. 7. If x 3 1 2x 2 1 1 5 ~x 1 1!~x 2 1 x 2 1! 1 r for every value of x, then r 5 . 8. The number of rational zeros of . f ~x! 5 ~x 2 2 2!~x 2 2 2x 1 3! is 9. The number of real roots of . ~x 2 2 2!~x 2 2 2x 1 3! 5 0 is 10. If x 37 2 2x 24 1 3x 2 2 5 is divided by x 1 1, then the remainder is . Develop Mastery Exercises 1–4 Locator Theorem Use the locator theorem to determine which half of the interval contains a zero of the function. 1. p~x! 5 x 3 2 3x 1 1; @22, 21# 2. f ~x! 5 2x 3 1 3x 2 2 x 2 2; @0,1# 3. g~x! 5 x 3 2 5x 2 1 5x 1 3; @21, 0# 4. p~x! 5 x 3 2 5x 2 1 7x 2 2; @2.5, 3# pg168 [V] 168 G2 5-36058 / HCG / Cannon & Elich Chapter 3 clb 11-22-95 MP1 Polynomial and Rational Functions Exercises 5–8 Division Algorithm Use division to find the polynomial part q~x! and remainder r when p~x! is divided by the given divisor. Write the result in the form p~x! 5 ~x 2 c!q~x! 1 r, and find p~c!. 5. p~x! 5 2x 3 1 3x 2 2 x 2 2; x 2 1 6. p~x! 5 2x 3 1 3x 2 2 x 2 2; x 1 2 7. p~x! 5 3x 4 1 x 3 2 2x 2 1 x 2 1; x 1 1 8. p~x! 5 x 3 2 3x 2 2 x 1 2; x 2 4 Exercises 9–10 Remainder Find the remainder when the polynomial is divided by x 2 c. 9. 4x 12 2 3x 8 1 5x 2 2 2x 1 3; c 5 21 10. x 10 2 64x 4 1 3; c 5 2 Exercises 11–14 Factor Theorem Use the factor theorem to find the value of k so that the given linear expression is a factor of the polynomial. 11. 2x 3 1 4x 2 1 kx 2 3; x 1 2 12. x 4 1 kx 2 1 kx 1 2; x 2 2 13. kx 3 1 3x 2 2 4kx 2 7; x 2 3 14. x 3 2 k 2x 1 ~k 1 1!; x 2 k Exercises 15–18 Remainder Theorem Use the remainder theorem to find the value of k so that when p~x! is divided by the linear expression you get the given remainder. 15. p~x! 5 2x 3 1 4x 2 1 kx 2 3; x 1 2; r 5 0 16. p~x! 5 2x 3 1 4x 2 1 kx 1 3; x 1 2; r 5 3 17. p~x! 5 x 3 1 kx 2 2 kx 1 8; x 2 2; r 5 1 18. p~x! 5 x 3 1 kx 2 2 kx 1 8; x 2 2; r 5 0 Exercises 19–24 Rational Zeros Theorem (a) Apply the Rational Zeros Theorem to list all of the possible rational zeros of f. If the theorem does not apply, explain why. (b) Use a calculator graph to help you eliminate some to the numbers listed in part (a). 19. f ~x! 5 6x 3 1 3x 2 2 2x 2 1 20. f ~x! 5 6x 3 2 2x 2 2 9x 1 3 21. f ~x! 5 2x 4 2 2x 3 2 6x 2 1 x 1 2 22. f ~x! 5 6x 3 2 x 2 2 13x 1 8 23. f ~x! 5 3x 3 2 1.5x 2 1 x 2 0.5 24. f ~x! 5 x 3 2 2x 2 1 Ï2x 2 2 Exercises 25–30 Exact Form Zeros (a) Find all zeros of f (including any complex numbers) in exact form. First look for rational zeros and express f ~x! in factored form (linear or quadratic factors). (b) Find the solution set for p~x! , 0. 25. f ~x! 5 x 3 2 4x 2 1 2x 2 8 26. f ~x! 5 4x 3 2 4x 2 2 19x 1 10 27. f ~x! 5 x 3 2 2.5x 2 2 7x 2 1.5 28. f ~x! 5 x 3 2 3.5x 2 1 0.5x 1 5 29. f ~x! 5 6x 4 2 13x 3 1 2x 2 2 4x 1 15 30. f ~x! 5 4x 4 2 4x 3 2 7x 2 1 4x 1 3 Exercises 31–32 Solving Polynomial Equations the solution set. 31. (a) 3x 2 2 12x 5 ~x 2 1!~x 2 2 4x! 3x 2 2 12x 5x21 (b) x 2 2 4x 32. (a) 3x 3 2 12x 5 ~x 1 2!~x 3 2 4x! 3x 3 2 12x 5 x 3 2 4x (b) x12 Find Exercises 33–34 Exact Form Roots (a) Find the roots in exact form. (Hint: The equation is quadratic in x 2. Use the quadratic formula to first find x 2.) (b) Get approximations (2 decimal places) to the answers in part (a). Graph as a check. 34. x 4 2 2x 2 2 1 5 0 33. x 4 2 4x 2 1 1 5 0 Exercises 35–38 Solution Set Find the solution set for (a) f ~x! 5 0, (b) f ~x 2 1! 5 0, (c) f ~x! # 0. (Hint: For part (c), first factor and get cut points.) 35. f ~x! 5 2x 3 2 3x 2 2 3x 1 2 36. f ~x! 5 x 4 2 2x 3 2 3x 2 1 4x 1 4 37. f ~x! 5 x 3 2 3x 1 2 38. f ~x! 5 4x 3 2 4x 2 2 19x 1 10 Exercises 39–42 Nonrational Numbers (a) Find a polynomial equation with integer coefficients having c as a root. (b) Explain why c is not a rational number. (Hint: See Example 7.) 39. c 5 Ï2 40. c 5 2 1 Ï5 3 3 42. c 5 2Ï31 1 41. c 5 Ï2 2 1 Exercises 43–46 Verbal to Formula (a) Find a formula (in expanded form) for a polynomial function satisfying the given conditions. (b) How many turning points does the graph have? 43. Degree 3; zeros are 22, 1, and 3; leading coefficient is 1. 44. Degree 3; zeros are 21, 2, and 4; leading coefficient is 22. 45. Degree 4; zeros are 21, 2, and a double zero at 1; graph of f contains the point (0, 22). 46. Degree 4; zeros are 0, 2, and ~x 2 2 2x 2 5! is a factor of f ~x!; graph contains the point (3, 6). Exercises 47–48 Zeros and Turning Points (a) How many real zeros does f have? (b) Find all rational zeros. (c) How many turning points does the graph of f have, if any? In what quadrants? 47. f ~x! 5 ~x 2 2 4!~x 2 2 8x 1 15! pg169 [R] G1 5-36058 / HCG / Cannon & Elich clb 11-22-95 MP1 3.2 Locating Zeros 169 48. f ~x! 5 ~2x 2 2 x 2 3!~x 2 1 2x 1 4! 49. Is there a polynomial function of degree 3 that has zeros at 22, 21, and 2, whose graph passes through the points (0, 4), (3,220)? Explain. 70. Repeat Exercise 69 for f ~x! 5 x 3 2 cx 1 16. 71. Cone in a Sphere A cone is inscribed in a sphere of radius 8 cm. See the diagrams showing cross sections. Let r denote the radius of the cone and h the height. Exercises 50–53 End Behavior and Local Minima The intercept points for a polynomial function f of degree 3 are given. (a) Draw a rough sketch and determine the end behavior. (b) Determine the coordinates (one decimal place) of the local minimum point. 50. ~22, 0!, ~1, 0!, ~3, 0!, ~0, 6! 51. ~22, 0!, ~1, 0!, ~3, 0!, ~0, 26! 52. ~24, 0!, ~22, 0!, ~1, 0!, ~0, 8! 53. ~25, 0!, ~21, 0!, ~1, 0!, ~0, 210! h Exercises 54–57 Local Maxima For the function in Exercises 50–53, give the coordinates (one decimal place) of the local maximum point. (a) Exercises 58–61 Your Choice Suppose the graph of a polynomial function f of degree 3 has a local maximum point at P and a local minimum point at Q. Draw a rough sketch (or sketches) of the graph of f and use it to describe any features such as the location of x-intercept points, end behavior, and any other properties. Write an equation for your function. 58. Both P and Q are in Quadrant I. 59. P is in Quadrant I and Q is in Quadrant IV. 60. P is in Quadrant II and Q is in Quadrant IV. 61. P is in Quadrant I and Q is in Quadrant III. Exercises 62–65 Bracketing Roots Find the smallest interval with integer endpoints @b, c# containing all the roots of the equation; that is, b is the largest integer smaller than all roots of the equation, and similarly for c. 62. 2x 3 2 3x 2 2 2x 1 3 5 0 63. x 3 2 3x 2 2 2x 1 4 5 0 64. x 4 2 2x 2 2 3 5 0 65. x 3 2 3x 2 2 2x 1 8 5 0 66. (a) For what number c is c a zero of f ~x! 5 2x 3 2 cx 2 1 ~3 2 c 2!x 2 6? (b) Using your value for c, draw a graph and verify that c is a zero of f. Exercises 67–68 Bracketing Zeros Find the smallest interval with integer endpoints @b, c# containing all the zeros of the function. See Exercises 62–65. 67. f ~x! 5 x 4 2 9x 2 1 6x 2 4 68. f ~x! 5 x 4 2 6x 2 1 3x 1 4 69. Explore Try integer values of c in f ~x! 5 x 3 2 cx 1 2 until you find one for which f has a repeated zero. Then determine the other zero. Justify your conclusions algebraically. r h 8 r (b) h r 8 (c) (a) Show that for both diagrams, r 2 5 16h 2 h 2. Express the volume V of the cone as a function of h. (b) Of all such possible cones, there is one that has the greatest volume. What are the radius and height (1 decimal place) giving maximum volume? What is the maximum volume? 72. Cone in a Sphere (Alternative Approach) Solve Exercise 71 by expressing V as a function of r. Explain why it is not necessary to consider the second diagram to find the maximum volume.