[PDF] 32 LOCATING ZEROS

101374_6Section_3_2.pdf pg158 [V] G2 5-36058 / HCG / Cannon & Elich clb 11-22-95 MP1

158Chapter 3 Polynomial and Rational Functions

3.2LOCATING 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

Ihad a good teacher for

In Section 3.1 we indicated that two of the important concerns for polynomials are freshman algebra. I think

locating zeros and local extrema. With calculator graphs we can make excellenthe was simultaneously the

football coach. Then I took approximations for both. On a graph, there is no obvious relation between zeros sophomore geometry. It and turning points, but in calculus we learn that every turning point of a polynomial was apparently thought functionfoccurs at a zero of another polynomial function called thederivative of that students couldn't learn

f. Thus the location of turning points also depends on locating zeros. We leave thegeometry in one year so

study of derivatives to calculus, but we devote this section to understanding more they had a second course about zeros of polynomial functions.in the junior year. The teacher in this second It would be nice to have something analogous to the quadratic formula for course didn't understand higher degree polynomials. For some polynomial functions, zeros can be expressed the subject and I did. I

in exact form using radicals and the ordinaryoperations of algebra, but in general,made a lot of trouble for

we must rely on approximations. See the two Historical Notes in this section. Some her. Saunders MacLaneofthetheoremsincludedinthissectionhelpusdeterminewhetherornotexactform 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 s ubsequent courses. Weneed 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 con®dent that there isn't some break or jump in the graph between two adjacent pixel columns? In connected mode, a graphing calculator connectsdiffer- ent points of the graph byvertical 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 Supposepis a polynomial function andaandbare numbers such thatp~a! andp~b!have opposite signs. The functionphas at least one zero betweena andb, or equivalently, the graph ofy5p~x!crosses thex-axis between ~a,0!and~b,0!.cEXAMPLE 1Using the locator theorem

Letp~x!52x

322x
2

23x11.

(a)Use the locator theorem to verify thatphas a zero between 0 and 1. (b)Using a decimal window, graphy5p~x!and trace to verify that the zero is located between 0.2 and 0.3. [-5, 5] by [-3.5, 3.5] (.2, .336) (.3, .026) p(x)=2x 3 -2x 2 -3x+1 pg159 [R] G1 5-36058 / HCG / Cannon & Elich jb 11-13-95 QC2

3.2 Locating Zeros159

Solution

(a)p~0!51andp~1!522, so there is a sign change and hence a zero between

0 and 1.

(b)Tracing along the graph ofp~x!in a decimal window, we can read the coordi- nates shown in Figure 11. Thusp~0.2!50.336 andp~0.3!520.026. There is a sign change between 0.2 and 0.3, so there is a zero in the interval. b 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 approxima- tion 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 zero- product principle. Unfortunately, in most cases where we need the zeros of a given polynomial function, there is no dependable procedure for ®nding 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 partqand a remainder r, where the remaindermust be smaller than the divisor, so we can divide one polynomial by another. The result of polynomial division is a polynomial partq~x!, and a remainderr~x!whose degree must be smaller than the degree of the divisor. In particular, when the divisor is a linear polynomial (of the formx2c), then the remainder is some numberr. 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

Ifp~x!is a polynomial of degree greater than zero, andd~x!is a polynomial, then dividingp~x!byd~x!yields unique polynomialsq~x!andr~x!,calledthe polynomial partandremainder,respectively, such that p~x!5d~x!´q~x!1r~x!, where the degree ofr~x!is smaller than the degree ofd~x!. Ifd~x!5x2c, then the remainder is a unique numberr, such that p~x!5~x2c!´q~x!1r. (1) To ®nd 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, thereisashortcutcalledsynthetic division.Synthetic division is stressed in tradi- tionalcoursesbecauseitisalsousedforseveraldifferentevaluationpurposes.With graphing calculators, however, the convenience of synthetic division does not jus- tify 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.

FIGURE 11

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160Chapter 3 Polynomial and Rational Functions

HISTORICAL NOTE

IS THERE A CUBIC FORMULA?

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 coef®cients. The steps are outlined in the following box. Synthetic division algorithm (for divisors of the formx2c)

To divide a polynomialp~x!of degreenbyx2c:

1.On the top line writec(change sign fromx2c), followed by all the

coef®cients ofp~x!in order of decreasing powers ofx, including any zero coef®cients.

2.Bring down the leading coef®cient, multiply byc, and add the product to

the next coef®cient to get the next entry on the bottom line. Repeat, multiplying the sum byc, and adding the product to the next coef®cient, and continue for all coef®cients ofp.

3.The ®rstnnumbers on the bottom line are the coef®cients ofq~x!,of

degreen21, and the last number is the remainderr. We illustrate the synthetic division algorithm with the problem from Exam- ple 2: Dividep~x!52x 3 22x
2

23x11byx11. You may want to compare

the coef®cients in the long division of Example 2 with the numbers in the synthetic division. Sincex115x2~21!, We write21onthetoplineattheleft,

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 who ®rst succeeded, and when, gets muddled by con¯icting claims. At least part of the credit belongs to

Scipione del Ferro (ca.1510). By

about 1540, Tartaglia had learned enough to win a public contest, solving 30 cubics in 30 days.

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

1ax1b50,

®rst calculate

A5 S a 3 D 3 1 S b 2 D 2 .

Cardan's solution is given by

x5 3

ÎÏA2b

22
3

ÎÏA1b

2.

Although Cardan is known for

developing the ®rst cubic formula, credit actually belongs to Tartaglia, pictured here. pg161 [R] G1 5-36058 / HCG / Cannon & Elich clb 11-22-95 MP1

3.2 Locating Zeros161

followed by the coef®cients ofp~x!in order of decreasing powers ofx: cfromx2cA21222231BCoef®cients ofp~x!

For each entry on22421B

middle line, multiply 22410
bottom entry by21, and add. q~x!52x 2

24x11r50

From the last line we read the coef®cients of the polynomial partq~x!, whenp~x! is divided byx11, and the remainderr. Writingp~x!in the form of the division algorithm, 2x 3 22x
2

23x115~x11!~2x

2

24x11!10.

cEXAMPLE 2Using the division algorithmVerifythat21isazeroofthe polynomial function from Example 1,p~x!52x 3 22x
2

23x11, and ®nd the

other two zeros in exact form.

Solution

Substituting21forx,p~21!50.

Dividingp~x!byx11 yields the following:

2x 2 24x11
x11 ) 2x 3 22x
2 23x11
2x 3 12x 2 24x
2 23x
24x
2 24x
x11 x11 0 Since the remainder is 0,p~x!can be written in factored form as 2x 3 22x
2

23x115~x11!~2x

2

24x11!.

By the zero-product principle, eitherx1150or2x

2

24x1150. Using the

quadratic formula for the second equation, we ®nd that the zeros ofpare21,

22Ï2

2 ,

21Ï2

2 . The second zero is about 0.29289, clearly the number we were

ªzeroing in onº in Example 1.

b c

EXAMPLE 3The division algorithm again

(a)Show thatx 2

14 is a factor of the polynomial

p~x!5x 4 2x 3 12x 2

24x28.(b)Find all zeros ofp~x!.

Solution

(a)We use long division. x 2 2x22 x 2 14 ) x 4 2x 3 12x 2 24x28
x 4 14x 2 2x 3 22x
2 24x
2x 3 24x
22x
2 28
22x
2 28
0 [-2, 3] by [-10, 2] p(x)=x 4 -x 3 +2x 2 -4x-8 pg162 [V] G2 5-36058 / HCG / Cannon & Elich clb 11-16-95 MP1

162Chapter 3 Polynomial and Rational Functions

By the division algorithm, since the remainder is 0, p~x!5~x 2 14!~x 2

2x22!,

and we have a factorization ofp~x!, thus reducing the problem of ®nding the zeros of a fourth polynomial functionpto solving two quadratic equations, x 2

1450, andx

2

2x2250.

(b)The four zeros ofp~x!are62i,21,and2.Withtworealzeros,thegraphof y5p~x!should cross thex-axis just twice. See Figure 12. b

Factor and Remainder Theorems

Examples 2 and 3 illustrate one of the keyconcepts in ®nding exact form zeros of polynomial functions. Finding a zero is equivalent to ®nding a factor, andonce we have factors, by the zero-product principle, the zeros ofpare the zeros of the factors. Further, since the degree ofqis smaller than the degree ofp,q~x!is called thereduced polynomial. In the case where the divisor is linear, the division algorithm provides two powerful theorems. Consider again Equation (1), p~x!5~x2c!q~x!1r. First, supposer50. Then~x2c!isafactor,p~x!5~x2c!q~x!. Conversely, if ~x2c!is a factor ofp~x!,thenp~x!5~x2c!q~x!,sormust be 0. Equation (1) is an identity, so we can replacexby any number and obtain a true statement. In particular, if we replacexbyc,weobtain p~c!5~c2c!q~c!1r50´q~c!1r501r5r. Thus the remainder always equals the value of thefunctionpat the numberc. Putting these observations together, we have the following.

Remainder and factor theorems

Whenp~x!is divided byx2c, the remainder isp~c!.

Whenp~x!is divided byx2c,thenx2cisafactorofp~x!if and only if p~c!50. The factor and remainder theorems give usways to ®nd a remainder without performing a lengthy division and can simplify many calculations. cEXAMPLE 4The remainder theorem (a)Find the remainder when the polynomialp~x!54x 15 15x 7 12x 4 13is divided byx11.

Strategy:By the remainder

(b)Find the value ofksuch that if the polynomialP~x!5x 3 1x 2

1kx24is

theorem, for(a)p~21!5r, divided byx12, then the remainder is 0. and for(b)P~22!50, from which we can solve

Solution

fork. (a)Following the strategy,p~21!54~21!15~21!12~1!13524. Hence, p~21!524, sor524. (b)EvaluatingPat22, we haveP~22!5281422k2452822k. Since the remainder whenP~x!is divided byx12is0,thenP~22!50. Thus,

2822k50, or22k58, ork524.

The desired function is

P~x!5x

3 1x 2

24x24.b

FIGURE 12

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3.2 Locating Zeros163

HISTORICAL NOTE

THERE IS NO ªQUINTIC FORMULAº

Clearing Fractions and Rational Zeros

Multiplying an equation by anonzero constant to clear fractions does not change the roots of the equation. For example, multiplying both sides of the equation x 3 17 2x 2 17 3x22 350,
by 6, we get an equation with integer coef®cients having the same roots, 6x 3 121x
2

114x2450.

If all coef®cients are integers, then the following theorem provides a complete list of all the rational numbers that can possibly be zeros.

Rational zeros theorem

Letpbe any polynomial function withinteger coef®cients.The only rational numbers that can possibly be zeros ofpare the numbers of the form r s , whereris a divisor of the constant term, andsis a divisor of the leading coef®cient. If none of these numbers is a zero, thenphas 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 technol- ogy, 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 calcula- tor can distinguish between rational and irrational numbers;every decimal is

Very shortly after discovery of

the general solution of thecubic equation (see ªIs There a Cubic

Formula?º), Ferrari (Italy, ca.

1545) derived a method for quartics

(polynomials of degree 4). For n52, 3, or 4, solutions for equa- tions of degreeninvolventh roots.

Whynotfordegree5?Nearlythree

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 forquintic equations.In Paris in 1829 another

18-year-old, Evariste Galois, took

the ®nal 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 coef®cients.

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. pg164 [V] G2 5-36058 / HCG / Cannon & Elich jb 11-13-95 QC2

164Chapter 3 Polynomial and Rational Functions

truncated (ªcut offº) to ®t the display capacity. Knowing what the rational possibil- itiesare,wecanuseacalculatortoverifythataparticularzeroisorisnotarational number. cEXAMPLE 5Rational possibilitiesUse the rational zeros theorem to list all possible zeros of the polynomial function. (a)P~x!5x 3 24x
2

1x26(b)R~x!5x

4 24x
3 1 14 9 x 2 1 44
9 x2 5 3

SolutionStrategy:(b)To get integer

coef®cients, multiply (a)ForP, begin by listing all possible numerators (factors of26) and denomina- throughby9andthenapply the rational zeros theorem. tors (factors of 1):

Zeros of 9Rare are the

Possible numerators:61, 2, 3, 6

same as the zeros ofR.

Possible denominators:61

The only possibilities for rational zeros ofPare the integers26,23,22,21,

1, 2, 3, and 6.

(b)Follow the strategy and ®nd all possible rational zeros ofS~x!59R~x!:

S~x!59x

4 236x
3 114x
2

144x215.

Possible numerators are factors of 15; denominators are factors of 9.

Possible numerators:61, 3, 5, 15

Possible denominators:61, 3, 9

The rational zeros theorem tells us thatS, and henceR, has only sixteen pos- sible rational zeros (in reduced form): 6 F

1, 3, 5, 15,

1 3 , 5 3 , 1 9 , 5 9 G .b Having listed lots of possibilities for rational zeros of the polynomialsPandR 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 6Finding rational zerosFind all rational zeros, if there are any,ofthepolynomialfunctionsPandRinExample 5.Approximateanyirrational zeros to two decimal place accuracy. (a)P~x!5x 3 24x
2

1x26(b)R~x!5x

4 24x
3 1 14 9 x 2 1 44
9 x2 5 3

Solution

(a)We don't know anything aboutPexcept that as acubic 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 atthelistofrationalzerosforPfrom Example 5, we see that the only positive rational possibilities are 1, 3, and 6. Therefore, the real zero of the polynomial Pcannot be a rational number. If we zoom into a very small box around the [-10, 10] by [-10, 10] (a) 4 8-4 -848 -4 -8 [3.8, 4.2] by [-0.1, 0.1] (b)

4 4.05 4.10 4.15

[-10, 10] by [-10, 10] (a)[-1.5, 4] by [-3.2, 3] (b) 123-1
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3.2 Locating Zeros165

x-interceptpointofthegraph(Figure 13b)andthentrace,we®ndthatthezero is very near 4.11. (b)For graphing purposes, we can choose either the polynomialR~x!, or the poly- nomialS~x!59R~x!with integer coef®cients, becauseRandShave the same zeros. If we were working by hand, most of us would chooseSto avoid fractions; for the calculator there is no difference, except thatRrequires a smaller window. The graph ofRis 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 ®nd zeros near21, 0.3, 1.6, and 3. Which,ifany,ofthesearerationalzeros?Thelistofrationalpossibilities(from

Example 5) includes21, 3,

1 3 ,and 5 3 , all of which are reasonable candidates, but are these actually zeros of the function? To decide, we need to evaluateRat each number (see the following Technology Tip). To calculator accuracy we ®nd that

R~21!50,R~3!50,R~

1 3 !50,R~ 5 3 !50.

We conclude thatRhas four rational zeros,21, 3,

1 3 ,and 5 3 .b

FIGURE 13

P~x!5x

3 24x
2 1x26

FIGURE 14

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166Chapter 3 Polynomial and Rational Functions

TECHNOLOGY TIPrFunction evaluation

Most calculators, when we trace along a curve, display coordinates. The y-coordinate is the calculated value corresponding to thex-coordinate of the pixel, but we have no way to specify a particularx-value unless our window happens to have it as a pixel. If our goal is to evaluateR~ 1 3 !,wedon'twantto settle forR(.324076113). We give here some suggestions fordifferent calculators, but there may be a more ef®cient way for your particular machine. The displayed value is the calculator's evaluation ofR~ 1 3 !, which may involve round-off error. For example, one of our calculators displays R~ 1 3 !,53E-13, meaning 0.0000000000003, and which in thiscontext we interpret as 0.

TI calculators:If you are graphing a function as

Y1, return to the home

screen, store the desired value in thex-register, and then enter

Y1. Thus, for

R~21!,21A

X Enter.ThencallupY1from theY-varsmenu (or on the TI-85,

2nd Alpha y1,andEnter.TheTI-82willevaluateY1(21)directly.

Casio calculators:The functionmust be entered on your

MEMlist, so type

in the function,

SHIFT MEM, F1(STO)andthenumber,say1forf

1 .To evaluatef 1 ~3!, EXIT and store 3 in thex-register,3AX EXE.ThenF2(RCL)1 EXE.

HP±38: Having entered a function as, say,

F1~X!, return to the home

screen, type

F1~1y3!,andENTER.

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 ` 424*A
`

3114/9*A

`

2144/9*A25/3'. ENTERtwice, so

youhaveanextracopyonthestack.Then,purple

NUMconverts to a number.

Foranothervalue,storeitin

A, Enter,andevaluateasanumber.

All these calculators except the TI-81 have some sort of

SOLVEroutine

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. Forinstance, if we know that some numbercis a zero of a polynomial function butcis not among the possibilities for rational zeros, then we can conclude thatcisnotarationalnumber,asinthenext example. cEXAMPLE 7Showing a number is not rational (a)Find a polynomial equation with integer coef®cients satis®ed by the number c5 3

Ï521.

(b)Use the rational zeros theorem to show thatcis not a rational number.

Solution

(a)The simplest polynomial equation satis®ed bycis x5 3

Ï521,

[-5, 5] by [-3.5, 3.5] on [-1, 0.5] < [2, `) 2x 3 -3x 2 -3x+2$0 (-1, 0)(2, 0)(.5, 0) pg167 [R] G1 5-36058 / HCG / Cannon & Elich jb 11-13-95 QC2

3.2 Locating Zeros167

but of course, this does not have integer coef®cients. To get integer coef®cients, we add 1, cube both sides, and simplify:@See the inside front cover for the expansion of~x11! 3 .] ~x11! 3 5~ 3

Ï5!

3 x 3 13x 2

13x1155

x 3 13x 2

13x2450.

Now we have a polynomial equation, one of whose solutions is the numberc.

Incidentally, a graph makes it clear thatp~x!5x

3 13x 2

13x24 has only

one real zero, which thus must be 3

Ï521.

(b)Since the leading coef®cient ofp~x!is 1, the only possible rational zeros ofp are6@1, 2, 4#. The numbercisazeroofpbut is not one of the possible rational zeros, so 3

Ï521 must be an irrational number.b

c EXAMPLE 8Solving an inequalityFind the solution set for 2x 3 23x
2

23x12$0.

Solution

A graph off~x!52x

3 23x
2

23x12isshowninFigure15.Itappearsfrom

the graph that the zeros are21, 1 2 , and 2, as we may verify, either by tracing in a decimal window, or by evaluating the function. The functionfis clearly positive between21and 1 2 , and wheneverx.2.Therefore,thesolutionsetfortheinequal- ity is given by

S5@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 functionp~x!54x

3

2xhas three real zeros.

2.The positive zero off~x!5x

3

23xis less than 1.73.

3.Forf~x!5x

3 21.6x
2

28.52x115.84, sincef~2!

andf~3!are positive, thenfcontains no zeros between

2 and 3.

4.Theequation2x

3 25x
2

14x2150 has no ration-

al roots.

5.The functionf~x!5~3x22!~x

2

22x24!has ex-

actly one real zero.

6.Whenx

3 22x
2

13x216 is divided byx23, then

the remainder is 2. Exercises 7±10 Fill in the blank so that the resulting statement is true. 7.Ifx 3 12x 2

115~x11!~x

2

1x21!1rfor ev-

ery value ofx,thenr5.

8.The number of rational zeros of

f~x!5~x 2 22!~x
2

22x13!is .

9.The number of real roots of

~x 2 22!~x
2

22x13!50is .

10.Ifx

37
22x
24
13x 2

25 is divided byx11, then the

remainder is .

Develop Mastery

Exercises 1±4Locator TheoremUse the locator theo- rem to determine which half of the interval contains a zero of the function.

1.p~x!5x

3

23x11;@22,21#

2.f~x!52x

3 13x 2

2x22;@0,1#

3.g~x!5x

3 25x
2

15x13;@21, 0#

4.p~x!5x

3 25x
2

17x22;@2.5, 3#

FIGURE 15

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168Chapter 3 Polynomial and Rational Functions

Exercises 5±8Division AlgorithmUse division to ®nd the polynomial part q~x!and remainder r when p~x!is di- vided by the given divisor. Write the result in the form p~x!5~x2c!q~x!1r,and ®nd p~c!.

5.p~x!52x

3 13x 2

2x22;x21

6.p~x!52x

3 13x 2

2x22;x12

7.p~x!53x

4 1x 3 22x
2

1x21;x11

8.p~x!5x

3 23x
2

2x12;x24

Exercises 9±10RemainderFind the remainder when

the polynomial is divided by x2c. 9.4x 12 23x
8 15x 2

22x13;c521

10.x 10 264x
4

13;c52

Exercises 11±14Factor TheoremUsethefactortheo-

rem to ®nd the value of k so that the given linear expression is a factor of the polynomial. 11.2x 3 14x 2

1kx23;x12

12.x 4 1kx 2

1kx12;x22

13.kx 3 13x 2

24kx27;x23

14.x 3 2k 2 x1~k11!;x2k Exercises 15±18Remainder TheoremUse the remain- der theorem to ®nd the value of k so that when p~x!isdivided by the linear expression you get the given remainder.

15.p~x!52x

3 14x 2

1kx23;x12;r50

16.p~x!52x

3 14x 2

1kx13;x12;r53

17.p~x!5x

3 1kx 2

2kx18;x22;r51

18.p~x!5x

3 1kx 2

2kx18;x22;r50

Exercises 19±24Rational Zeros Theorem (a)Apply

the Rational Zeros Theorem to list all of the possible ratio- nal zeros of f. If the theorem does not apply, explain why. (b)Useacalculatorgraphtohelpyoueliminatesometothe numbers listed in part (a).

19.f~x!56x

3 13x 2 22x21

20.f~x!56x

3 22x
2 29x13

21.f~x!52x

4 22x
3 26x
2 1x12

22.f~x!56x

3 2x 2

213x18

23.f~x!53x

3 21.5x
2

1x20.5

24.f~x!5x

3 22x
2

1Ï2x22

Exercises 25±30Exact 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!5x

3 24x
2 12x28

26.f~x!54x

3 24x
2

219x110

27.f~x!5x

3 22.5x
2

27x21.5

28.f~x!5x

3 23.5x
2

10.5x15

29.f~x!56x

4 213x
3 12x 2

24x115

30.f~x!54x

4 24x
3 27x
2 14x13 Exercises 31±32Solving Polynomial EquationsFind the solution set.

31. (a)3x

2

212x5~x21!~x

2 24x!
(b) 3x 2 212x
x 2

24x5x21

32. (a)3x

3

212x5~x12!~x

3 24x!
(b) 3x 3 212x
x125x 3 24x
Exercises 33±34Exact Form Roots (a)Find the roots in exact form. (Hint: The equation is quadratic in x 2 .Use the quadratic formula to ®rst ®nd x 2 .)(b)Get approxima- tions (2 decimal places) to the answers in part (a). Graph as a check. 33.x
4 24x
2

115034.x

4 22x
2 2150
Exercises 35±38Solution SetFind the solution set for (a)f~x!50,(b)f~x21!50,(c)f~x!#0.(Hint: For part (c), ®rst factor and get cut points.)

35.f~x!52x

3 23x
2 23x12

36.f~x!5x

4 22x
3 23x
2 14x14

37.f~x!5x

3 23x12

38.f~x!54x

3 24x
2

219x110

Exercises 39±42Nonrational Numbers (a)Find a

polynomial equation with integer coef®cients having c as a root.(b)Explain why c is not a rational number. (Hint: See

Example 7.)

39.c5Ï240.c521Ï5

41.c5
3

Ï22142.c52

3

Ï311

Exercises43±46VerbaltoFormula (a)Findaformula

(in expanded form) for a polynomial function satisfying the given conditions.(b)How many turning points does the graph have?

43.Degree 3; zeros are22, 1, and 3; leading coef®cient

is 1.

44.Degree 3; zeros are21, 2, and 4; leading coef®cient is

22.

45.Degree4;zerosare21,2,andadoublezeroat1;graph

offcontains the point (0,22).

46.Degree 4; zeros are 0, 2, and~x

2

22x25!isafactor

off~x!; graph contains the point (3, 6).

Exercises 47±48Zeros 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 24!~x
2

28x115!

(a)h r (b) h r 8 (c) h r 8 pg169 [R] G1 5-36058 / HCG / Cannon & Elich clb 11-22-95 MP1

3.2 Locating Zeros169

48.f~x!5~2x

2

2x23!~x

2

12x14!

49.Isthereapolynomialfunctionofdegree3thathaszeros

at22,21,and2,whosegraphpassesthroughthe points (0, 4), (3,220)? Explain. Exercises 50±53End Behavior and Local MinimaThe intercept points for a polynomial function f of degree 3 are given.(a)Draw a rough sketch and determine the end be- havior.(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!

Exercises 54±57Local MaximaFor the function in Ex- ercises 50±53, give the coordinates (one decimal place) of the local maximum point. Exercises 58±61Your ChoiceSuppose 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.BothPandQare in Quadrant I.

59.Pis in Quadrant I andQis in Quadrant IV.

60.Pis in Quadrant II andQis in Quadrant IV.

61.Pis in Quadrant I andQis in Quadrant III.

Exercises62±65Bracketing RootsFind the smallest in- terval 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 23x
2

22x1350

63.x
3 23x
2

22x1450

64.x
4 22x
2 2350
65.x
3 23x
2

22x1850

66. (a)For what numbercisca zero of

f~x!52x 3 2cx 2 1~32c 2 !x26? (b)Using your value forc,drawagraphandverifythat cisazerooff. Exercises67±68Bracketing ZerosFind the smallest in- terval with integer endpoints@b,c#containing all the zeros of the function.See Exercises 62±65.

67.f~x!5x

4 29x
2 16x24

68.f~x!5x

4 26x
2 13x14

69.ExploreTry integer values ofcinf~x!5x

3 2 cx12 until you ®nd one for whichfhas a repeated zero. Then determine the other zero. Justify your con- clusions algebraically.70.Repeat Exercise 69 forf~x!5x 3

2cx116.

71.Cone in a SphereA cone isinscribed in a sphere of

radius 8 cm. See the diagrams showing cross sections.

Letrdenote the radius of thecone andhthe height.

(a)Show that for both diagrams,r 2

516h2h

2 .Ex- press the volumeVof the cone as a function ofh. (b)Of all such possiblecones, there is one that has the greatest volume. What are the radius and height (1 decimal place) giving maximum volume? What is themaximumvolume?

72.Cone in a Sphere (Alternative Approach)Solve Exer-

cise 71 by expressingVas a function ofr.Explainwhy it is not necessary to consider the seconddiagram to ®nd the maximum volume.