Maths Quest Methods year 12

VCE mATHEmATiCS UNiTS 3 & 4

mATHS QUEST

12

mathematical

methods CAS

Contents

Contents v

introduction viii

Acknowledgements xi

T i - N S P i R E 2 . 0 E D i T i O N

VCE mATHEmATiCS

UNiTS 3 & 4

CONTRiBUTiNG AUTHORS

RUTH BAKOGiANiS | ANDREW mENTLiKOWSKi | mARK BARNES | KYLiE BOUCHER JENNY WATSON | CAROLiNE DENNEY | SONJA STAmBULiC | ELENA iAmPOLSKY

ROSS ALLEN | ROBERT CAHN | RODNEY EBBAGE

BRiAN HODGSON NiCOLAOS KARANiKOLAS

BEVERLY LANGSFORD-WiLLiNG mARK DUNCAN TRACY HERFT

LiBBY KEmPTON JENNiFER NOLAN GEOFF PHiLLiPS

mATHS QUEST

₁₂

mathematical

methods CAS

Screenshots from TI-Nspire CAS calculator reproduced with permission of Texas Instruments

First published and revised 2010 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10/12pt Times LT

© John Wiley & Sons Australia, Ltd 2010 The moral rights of the authors have been asserted. National Library of Australia

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Title: Maths quest 12 mathematical methods CAS: TI-Nspire 2.0 edition / Brian Hodgson . . . [et al.]. ISBN: 978 1 7424 6465 7 (student pbk.)

978 1 7424 6466 4 (student ebook) 978 1 7424 6468 8 (teacher pbk.) 978 1 7424 6469 5 (teacher ebook)

Notes: Includes index.

Target Audience: For secondary school age. Subjects: Mathematics — Textbooks. Other Authors/

Contributors: Hodgson, Brian. Dewey Number: 510

Contents

Introduction viii About eBookPLUS x Acknowledgements xi

CHAPTER 1

Graphs and polynomials 1

1a The binomial theorem 1 Exercise 1A 6 1B Polynomials 6 Exercise 1B 10 1C Division of polynomials 11 Exercise 1C 15 1D Linear graphs 15 Exercise 1D 20 1E Quadratic graphs 22 Exercise 1E 28 1F Cubic graphs 29 Exercise 1F 34 1G Quartic graphs 38 Exercise 1G 45 Summary 47 Chapter review 50 eBookPLUS activities 56 CHAPTER 2

Functions and transformations 57

2a Transformations and the parabola 57 Exercise 2A 63

2B The cubic function in power form 65 Exercise 2B 69

2C The power function (the hyperbola) 71 Exercise 2C 76

2D The power function (the truncus) 78 Exercise 2D 84

2E The square root function in power form 86 Exercise 2E 91

2F The absolute value function 92 Exercise 2F 95

2G Transformations with matrices 96 Exercise 2G 101

2h Sum, difference and product functions 103 Exercise 2H 106

2I Composite functions and functional equations 108 Exercise 2I 110 2J Modelling 111 Exercise 2J 117 Summary 120 Chapter review 123 eBookPLUS activities 129 ExAm PRACTiCE 1 Based on Chapters 1–2 130 CHAPTER 3

Exponential and logarithmic

equations 132

3a The index laws 132 Exercise 3A 136 3B Logarithm laws 138

Exercise 3B 141

3C Exponential equations 142 Exercise 3C 146

3D Logarithmic equations using any base 147 Exercise 3D 149

3E Exponential equations (base e) 151 Exercise 3E 153

3F Equations with natural (base e) logarithms 154 Exercise 3F 155 3G Inverses 156 Exercise 3G 158 3h Literal equations 159 Exercise 3H 162

3I Exponential and logarithmic modelling 162 Exercise 3I 164

Summary 167

Chapter review 168

eBookPLUS activities 171

CHAPTER 4

Exponential and logarithmic graphs 172

4a Graphs of exponential functions with any base 172

Exercise 4A 181

4B Logarithmic graphs to any base 182 Exercise 4B 190

4C Graphs of exponential functions with base e 191

4D Logarithmic graphs to base e 197 Exercise 4D 201

4E Finding equations for graphs of exponential and logarithmic functions 202

Exercise 4E 205

4F Addition of ordinates 206 Exercise 4F 211

4G Exponential and logarithmic functions with absolute values 212

Exercise 4G 215

4h Exponential and logarithmic modelling using graphs 216 Exercise 4H 217 Summary 220 Chapter review 222 eBookPLUS activities 228 CHAPTER 5

inverse functions 229

5a Relations and their inverses 229 Exercise 5A 233

5B Functions and their inverses 234 Exercise 5B 239 5C Inverse functions 240 Exercise 5C 243 5D Restricting functions 245 Exercise 5D 249 Summary 252 Chapter review 253 eBookPLUS activities 257 CHAPTER 6

Circular (trigonometric) functions 258

6a Revision of radians and the unit circle 258 Exercise 6A 263

6B Symmetry and exact values 264 Exercise 6B 271

6C Trigonometric equations 273 Exercise 6C 281

6D Trigonometric graphs 282 Exercise 6D 288

6E Graphs of the tangent function 291 Exercise 6E 295

6F Finding equations of trigonometric graphs 295 Exercise 6F 297 6G Trigonometric modelling 299 Exercise 6G 301 6h Further graphs 303 Exercise 6H 310

6I Trigonometric functions with an increasing trend 311 Exercise 6I 313 Summary 314 Chapter review 317 eBookPLUS activities 321 ExAm PRACTiCE 2 Based on Chapters 1–6 322 CHAPTER 7

Differentiation 324

7a Review — gradient and rates of change 324 Exercise 7A 328

7B Limits and differentiation from first principles 332

Exercise 7B 336

7C The derivative of xn ₃₃₈ Exercise 7C 340

7D The chain rule 341 Exercise 7D 343

7E The derivative of ex ₃₄₄ Exercise 7E 347

7F The derivative of loge (x) 347 Exercise 7F 349

7G The derivatives of sin (x), cos (x) and tan (x) 351

Exercise 7G 354 7h The product rule 355

Exercise 7H 356 7I The quotient rule 357

Exercise 7I 359

7J Mixed problems on differentiation 360 Exercise 7J 364 Summary 366 Chapter review 368 eBookPLUS activities 372 CHAPTER 8

Applications of differentiation 373

8a Equations of tangents and normals 373 Exercise 8A 375

8B Sketching curves 376 Exercise 8B 384

8C Maximum and minimum problems when the function is known 386 Exercise 8C 389

8D Maximum and minimum problems when the function is unknown 390 Exercise 8D 394

8E Rates of change 396 Exercise 8E 398 8F Related rates 401 Exercise 8F 404 8G Linear approximation 404 Exercise 8G 406 Summary 408 Chapter review 410 eBookPLUS activities 415 CHAPTER 9

integration 416

9B Integration of e x_{, sin (x) and cos (x) 426} Exercise 9B 428

9C Integration by recognition 430 Exercise 9C 433

9D Approximating areas enclosed by functions 435

Exercise 9D 438

9E The fundamental theorem of integral calculus 441 Exercise 9E 446 9F Signed areas 448 Exercise 9F 452 9G Further areas 455 Exercise 9G 458

9h Areas between two curves 460 Exercise 9H 464

9I Average value of a function 466 Exercise 9I 468

9J Further applications of integration 468 Exercise 9J 471 Summary 474 Chapter review 476 eBookPLUS activities 481 ExAm PRACTiCE 3 Based on Chapters 1–9 482 CHAPTER 10

Discrete random variables 484

10a Probability revision 484 Exercise 10A 498

10B Discrete random variables 501 Exercise 10B 507

10C Measures of centre of discrete random distributions 510

Exercise 10C 516

10D Measures of variability of discrete random distributions 518 Exercise 10D 526 Summary 529 Chapter review 531 eBookPLUS activities 536 CHAPTER 11

The binomial distribution 537

11a The binomial distribution 537 Exercise 11A 548

11B Problems involving the binomial distribution for multiple probabilities 552

Exercise 11B 557

11C Markov chains and transition matrices 559 Exercise 11C 571

11D Expected value, variance and standard deviation of the binomial distribution 573 Exercise 11D 577 Summary 581 Chapter review 583 eBookPLUS activities 588 CHAPTER 12

Continuous distributions 589

12a Continuous random variables 589 Exercise 12A 594

12B Using a probability density function to find probabilities of continuous random variables 595

Exercise 12B 602

12C Measures of central tendency and spread 605

Exercise 12C 611

12D Applications to problem solving 613 Exercise 12D 616

12E The normal distribution 619 Exercise 12E 623

12F The standard normal distribution 626 Exercise 12F 633

12G The inverse cumulative normal distribution 636 Exercise 12G 640 Summary 643 Chapter review 646 eBookPLUS activities 651 ExAm PRACTiCE 4 Based on Chapters 1–12 652 Answers 654 Index 730

introduction

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AreAS oF STudy

Graphs of polynomial functions including •

axis intercepts, stationary points and points of infl ection, domain (including maximal domain), range and symmetry

Review of algebra of polynomials, solution of •

polynomial equations with real coeffi cients of degree n having up to and including n real solutions

1A The binomial theorem

1B Polynomials 1C Division of polynomials 1D Linear graphs 1E Quadratic graphs 1F Cubic graphs 1G Quartic graphs

1

Graphs and

polynomials

The binomial theorem

In Maths Quest 11 Mathematical Methods CAS we learned the following binomial expansions: (x + a)2 _{= x}2 _{+ 2xa + a}2

(x + a)3_{= x}3_{+ 3x}2_{a + 3xa}2_{+ a}3

These are called binomial expansions because the expressions in the brackets contain two terms, ‘bi’ meaning 2.

By continuing to multiply successively by a further (x + a), the following expansions would be obtained:

(x + a)4 _{= (x}3 _{+ 3x}2_{a + 3xa}2 _{+ a}3_{)(x + a)} = x4 _{+ 4x}3_{a + 6x}2_a2 _{+ 4xa}3 _{+ a}3 (x + a)5 _{= (x}4 _{+ 4x}3_{a + 6x}2_a2 _{+ 4xa}3 _{+ a}3_{)(x + a)}

= x5 _{+ 5x}4_{a + 10x}3_a2 _{+ 10x}2_a3 _{+ 5xa}4 _{+ a}5

The coeffi cients associated with each term can be arranged in a triangular shape as shown:

(x + a)0 ₁ (x + a)1 _{1 1} (x + a)2 _{1 2 1} (x + a)3 _{1 3 3 1} (x + a)4 _{1 4 6 4 1} (x + a)5 _{1 5 10 10 5 1}

1A

Notes

1. The first and last numbers of each row are 1.

2. Each other number is the sum of the two numbers immediately above it.

This triangle is known as Pascal’s triangle. Each number can also be obtained using combinations, as follows. Row 0 0 0     1 1 0     1 1     2 2 0     2 1     2 2     3 3 0     3 1     3 2     3 3     4 4 0     4 1     4 2     4 3     4 4     Note: n_r nCr _{n r r}n    = =

(

)

r is another way of writing

n r    .

For example, the expansion of (x + a)6 _{can be written using combinations and} then evaluated: (x a+ ) = x a x a x    +     +     6 6 6 0 5 1 4 0 6 1 6 2 aa x a x a x a 2 6 3 3 2 4 1 5 3 6 4 6 5 +    +     +     + 66 6 6 15 20 15 0 6 6 5 4 2 3 3 2 4     = + + + + + x a x x a x a x a x a 66_xa5_+a6

Now the binomial theorem can be formally stated. (ax b₊ )n₌ n ( )ax bn ₊ n ( )ax n b ₊. 0 1 0 1 1         −− .. .+ ( ) + ( ) n n ax b n n ax b n n −−         −− 1 1 1 0 Notes

1. The indices always sum to n, that is, the powers of (ax) and b sum to n. 2. The power of ax decreases from left to right while the power of b increases. 3. The number of terms in the expansion is always n + 1.

4. The (r + 1)th term is n r     (ax) n − r _br_.

The binomial theorem can also be stated using summation notation:

(ax b) n ( ) r ax b n n r r r n + =     − =

∑

∑

Worked exAMple 1

Use the binomial theorem to expand (2_x−−3)4_.

ThINk WrITe

Method 1: Technology-free

1 Complete the binomial theorem expansion where ax is the 1st term, b is the 2nd term and n is the index, using the appropriate row of Pascal’s triangle to assist.

(2 3) 4 ( ) ( ) ( ) ( 0 2 3 4 1 2 3 4 4 0 3 x− = x x    +     − − ₎₎1 +            +             − ₊ − − ₊ − −  −   −  −    −  −   −  −   4 2 2 3 4 3 2 3 − ₂ −₃ − − 2 − 2 − 2 − 2 − 2 2 32 2 −−−−3 ₂₂₂₂ 111 −−−−₃₃₃₃333 ( )2 3 ( )₂2_222222222xxx 22222222_{( )( )( )( )( )( )}−−−−−−−−−−−−−−−−−−−−−−−−−−₃₃3_3333333322222222 ₊₊ _{( )( )( )( )( )22222222222222xxx}111111_{( )( )( )( )( )}−−−−−−−−−−−−−−−−−−−−−−−−−−_{33333333333333}333333  x x  x __ x x _ x  x x   x  x  x x  x __ x x _ x  x x  2 2 x 2 2 x 2x 3 x 2 3 2 2 32 2x 3 x 2 2 32 2 3 2 3 ( )22x 33 x ( )2₂2₂₂2₂₂2₂2_22xx_xxxx 22222222_{( )( )}( )_{( )( )}( )( )( )₃₃₃3₃3₃333₃₃₃22222222 _{( )( )( )( )2222}x_xxxx_{x 3333} +            4 4 2 3 0 4 2 0 34 2 3 ( )2 3 ( )₂₂₂₂₂₂2₂₂₂22 000000( )_{( )( )}( )_{( )}−−−−−−−−₃33_33333333₃444444 ( )22x 33 ( )2 3

2 Evaluate the combinations and the powers. = 1(16x4_{) + 4(8x}3₎₍−_{3) + 6(4x}2₎₍₉₎

+ 4(2x)(−_{27) + 1(81)}

3 Simplify. = 16x4 _{− 96x}3_{+ 216x}2 _{− 216x + 81}

Method 2: Technology-enabled 1

2 Write the result. (2x − 3)4_{= 16x}4_{− 96x}3_{+ 216x}2_{− 216x + 81}

Worked exAMple 2

Expand the binomial expression 2 2 5 x +x     . ThINk WrITe

1 Complete the binomial expansion where ax

x

= 2₂, b = x and n = 5, using row 5 of Pascal’s triangle to assist. 2 2 5 2 10 2 2 5 2 5 2 4 2 x +x x x x x    =   +   +   3 2 x + 10_ 2₂_ + 5_ 2_ + 2 3 2 4 5 x x x x x

2 Evaluate the powers. == ++ 

      +        32 5 16 10 8 10 = ₁₀ + = + _{8 6}  8_ 6 8_ + 6 8 + _ 6 8 _ 6 8 10_ 6 8 10 6 2 x x x x x10 x x10 x8888 xx _xx6666 x +    +   + 10 4₄ 3 5 2 2 4 5 x x x x x 3 Simplify. = 32₁₀ +80₇ +80₄ +40+10 2 + 5 x x x x x x

On a Calculator page, press: •  MENu b

•  3:Algebra 3 •  2:Expand 2

Complete the entry line as: expand((2x − 3)4₎

Worked exAMple 3

State the coeffi cient of i x2 _{and ii x}4 _{in (3 2−− x , without the use of technology.)}8

ThINk WrITe

i 1 The powers of the 1st term decrease and the powers of the 2nd term increase 0, 1, 2, . . . use this to fi nd which term gives a power of x2_.

i x0_{, x}1_{, x}2

The third term gives a power of x2_.

2 Find the appropriate term by using the binomial theorem. Third term = 8 2     3 6 ₍−_2x)2

3 _{Evaluate the term.} = 28 × 729 × 4x2

= 81 648x2

4 State the coeffi cient. The coeffi cient of x2 _{is 81 648.} ii 1 _{Find which term gives a power of x}4_{. ii x}0_{, x}1_{, x}2_{, x}3_{, x}4

The fi fth term gives a power of x4_.

2 Evaluate the term. Fifth term = 8

4     3 4 ₍−_2x)4 = 70 × 81 × 16x4 = 90 720x4

3 State the coeffi cient. The coeffi cient of the fi fth term is 90 720.

Worked exAMple 4

Find the fourth term in the expansion of (x−− 2y)5_.

ThINk WrITe

1 _{Find the fourth term by using the binomial} theorem. Fourth term = 5 3     x 2_(2y)3

2 Evaluate the term. = 10 × x2× 8y3

= 80x2_y3

Worked exAMple 5

Find and evaluate the term that is independent of x in the expansion of x x 3 2 5 1 +    . ThINk WrITe

1 _{Find how the powers of x are generated} in the expansion from left to right.

Powers of x are (x3₎5 _{= x}15_{, (x}3₎4 1 2 x    =x10, (x3₎3 1 2 2 x    =x5, (x3)2 1 2 3 x    =x0 . . . that is, x15_{, x}10_{, x}5_{, x}0_. eBookplus eBookplus Tutorial int-0516 Worked example 3

2 Find the required term. The fourth term is independent of x.

3 Evaluate. Fourth term = 5

3 1 3 2 2 3        ( )x x = 10 6 1 6 x x         = 10

4 State the solution. The term that is independent of x is the fourth term, 10.

Worked exAMple 6

Find the coefficient of y4 _{in the expansion of (y++3) (}3 _2−−y)5_.

ThINk WrITe

1 _y4 _{terms will result when multiplying from the} first and second brackets respectively: terms 1 and 2, terms 2 and 3, terms 3 and 4 and terms 4 and 5.

2 Write down the sum of these 4 products, using Pascal’s triangle to assist.

y4 _{terms = y}3_[5(2)4₍−_{y)] + 3y}2_(3)[10(2)3₍−_y)2_]

+ 3y(3)2_[10(2)2₍−_y)3_{] + 3}3_[5(2)(−_y)4_]

3 Evaluate. = −_80y4_{+ 720y}4 _{− 1080y}4_{+ 270y}4

= −_170y4

4 State the solution. The coefficient of y4 _is −_170.

Pascal’s triangle: 1. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Binomial theorem: 2. (ax b₊ )n₌n ( )ax bn n ( )ax n b ..    +     − + 0 1 0 1 _{..+ ( ) ( )} −     +     − n n ax b n n ax b n n 1 1 0 Notes The powers of (

1. ax) and b sum to n.

There are

2. n + 1 terms in the expansion. The ( 3. r + 1)th term is n r ax n r rb    ( ) − . reMeMBer

The binomial theorem

1 We 1 use the binomial theorem to expand each of the following. a (x + 3)2 _{b (x + 4)}5 _{c (x − 1)}8

d (2x + 3)4 _{e (7 − x)}4 f (2 − 3x)5 2 We2 Expand each of the following binomial expansions.

a x x +     1 3 _b 3 2 7 x x −     c x2 x 6 3 +     d 3 2 2 5 x − x     3 We3 State the coefficient of i x2 _{ii x}3 _{and iii x}4 _{in each of the following.}

a (x − 7)3 _{b (2x + 1)}5 _c 2 ₃ 5 x+ x     d x2 x 6 3 −     e 7 3 2 6 x x +     4 MC The coefficient of x3 _{in 3} 2 5 3 x x −     is: A −_{135 B} −_{45 C} −_{75 D 45 E 135} 5 MC Which of the following does not have an x5 _{term when expanded?}

A (x + 6)8 _{B 3x2} 1 7 x −    C 6 5 8 x x +    D (8 − 3x) 5 _{E 2} 1 2 8 x x −    6 MC If x x ax bx cx d e x f x 3 2 5 15 10 5 5 10 2 +    = + + + + + , then a + b + c + d + e + f equals: A 15 B 31 C 63 D 243 E 127 7 MC Which one of the following expressions is not equal to (2x − 3)4?

A (3 − 2x)4 _{B (2x − 3)(2x − 3)}3 _C 2 3 3 2 6 2 x x −

(

)

(

)

9 Find the third term in the expansion of 3 4 9 −    x

, assuming ascending powers of x. 10 We5 Find and evaluate the term that is independent of x in the expansion of 3 2₂

6 x x +    . 11 Find and evaluate the term independent of x in the expansion of x

x 2 3 5 4 −    . 12 Find and evaluate the term that is independent of x in the expansion of x

x 2 2 4 3 +    . 13 We6 Find the coefficient of p4 in the expansion of (p + 3)5 (2p − 5).

14 In the expansion of (2a − 1)n_{, the coefficient of the second term is} −_{192. Find the value of n.}

polynomials

A polynomial in x is an expression that consists of terms which have non-negative integer powers of x only. P(x) is a polynomial in x if: P(x) = an xn+ an − 1 xn − 1+ . . . + a2 x2+ a1 x + a0 exerCISe

1A

1B

where n is the degree (or highest power) of the polynomial and is a non-negative integer. The values of an, an − 1, . . ., a2, a1 and a0 are called the coefficients of their respective power of x terms.

Worked exAMple 7

Which of the following expressions are not polynomials?

a x6₋₋₄x4₊₊₂x3₊₊₇x _{b x}92 _+x3_−x2_+6x−5 c 7 3_{−− xy++}4_x2_−−x3_{++ x d 8 2++ x−−3x}2_++9x3_−−x4 e 3 2 2 2 x x −     ThINk WrITe

1 _a and d are polynomials because they are expressions with non-negative integer powers of x only.

2 _{b is not a polynomial as it has a power of} 9 2, which

is not an integer.

3 _c is not a polynomial as it has a power of

1

2

(

)

has one term, −_{3xy, which is not a power} of x only.

b, c and e are not polynomials.

4 _e is not a polynomial because 2₂ 2 2

x = x

−

and so has a power that is not a positive integer.

Polynomials can be added and subtracted by collecting like terms.

Worked exAMple 8

Given that P x( )== −−6 2x++3x2++x Q x4, ( )==x5−−2x4++x2−−5x−−2_{and (( )}R _{x ==x}2_{−−4, find:} a P x( )++Q x( )

b P x( )−−R x( ).

ThINk WrITe/dISplAy

Method 1: Technology-free

a 1 Add the polynomials. a P(x) + Q(x) = 6 − 2x + 3x2+ x4+ x5 − 2x4 + x2 _{− 5x − 2}

2 Collect like terms. = x5 _{− x}4 _{+ 4x}2 _{− 7x + 4}

b 1 Subtract the polynomials. b P(x) − R(x) = 6 − 2x + 3x2_{+ x}4 _{− (x}2 _{− 4)}

2 Remove brackets. = 6 − 2x + 3x2_{+ x}4 _{− x}2_{+ 4}

Method 2: Technology-enabled

a 1

2

b

evaluating polynomials

A value for a polynomial, P(x), can be found for a particular value of x by simply substituting the given value of x into the polynomial expression and evaluating. That is, polynomial functions are evaluated in the same way as any function.

On a Calculator page, define the polynomials P(x), Q(x) and R(x). To do this, press:

•  MENu b •  1:Actions 1 •  1:Define 1

Complete the entry lines as: Define p(x) = 6 − 2x + 3x2_{+ x}4 Define q(x) = x5_{− 2x}4_{+ x}2 _{− 5x − 2} Define r(x) = x2_{− 4}

Press ENTER · after each entry.

To calculate P(x) + Q(x), complete the entry line as:

p(x) + q(x) Press ENTER ··

To calculate P(x) − R(x), complete the entry line as:

p(x) − r(x)

Worked exAMple 9

For the polynomial P x( )==2x4−−x3++5x2−−6x++4_{, find:}

a its degree b P(1) c P(−_2).

ThINk WrITe

a The degree of the polynomial is the highest power of x.

a The degree of P(x) is 4. b 1 Substitute the given value of x into

the polynomial expression.

b P(1) = 2(1)4 _{− (1)}3_{+ 5(1)}2 _{− 6(1) + 4}

2 Evaluate. = 2 − 1 + 5 − 6 + 4 = 4

c 1 Substitute the given value of x into the polynomial expression.

c P(−_{2) = 2(}−₂₎4 _{− (}−₂₎3 _{+ 5(}−₂₎2 _{− 6(}−_{2) + 4}

2 _{Evaluate. = 32 + 8 + 20 + 12 + 4 = 76}

Worked exAMple 10

If p x( )==ax5++x4−−₃x3++bx−−₅, ( )p−−₁ ==−−_{5 and (2)p} ==−−_{65, find the values of a and b.}

ThINk WrITe/dISplAy

Method 1: Technology-free

1 Substitute a given value of x into the polynomial and equate it to the given answer.

P(−_{1) = a(}−₁₎5 _{+ (}−₁₎4 _{− 3(}−₁₎3 _{+ b(}−_{1) − 5}

= −₅

2 _{Simplify the equation.} −_{a + 1 + 3 − b − 5 =} −₅

−_{a + 4 − b = 0}

3 Make b the subject of the equation and call this equation [1].

b = 4 − a [1]

4 _{Substitute a given value of x into the} polynomial and equate it to the given answer.

P(2) = a(2)5 _{+ (2)}4 _{− 3(2)}3 _{+ b(2) − 5}

= −₆₅

5 Simplify the equation. 32a + 16 − 24 + 2b − 5 = −₆₅

32a + 2b − 13 = −₆₅

32a + 2b = −_{52 [2]}

6 Substitute [1] into [2]. Substituting b = 4 − a:

32a + 2(4 − a) = −₅₂

7 Solve this equation for a. 32a + 8 − 2a = −₅₂

30a = −₆₀ a = −₂

8 _{Substitute the value of a into equation [1]. Substituting a =} −_{2 into equation [1]:} b = 4 − −₂

9 Find the value of b. = 6

Method 2: Technology-enabled 1

2 Write the answer. Given p(x) = ax5_{+ x}4_{− 3x}3_{+ bx − 5 and solving} p(−_{1) =} −_{5 and p(2) =} −_{65 gives a =} −_{2 and b = 6.}

If

1. P(x) = a nx n+ an − 1xn − 1+ . . . + a2x2+ a1x + a0 and n is a non-negative integer then P(x) is a polynomial of degree n and a_n, a_{n − 1}, . . . a₂, a₁ are called coeffi cients and ∈ R. A polynomial

2. P(x) is evaluated in the same way as any function. reMeMBer

polynomials

1 We 7 Which of the following are not polynomial expressions?

i x3 _{− 2x ii x}4_{+ 3x}2 _{− 2x + x iii x}7 _{+ 3x}6 _{− 2xy + 5x} iv 3x8 _{− 2x}5_{+ x}2 _{− 7 v 4x}6 _{− x}3_{+ 2x − 3 vi 2x}5 _x4 _x3 _x2 _3x 2

x

+ − + + −

2 We8 Given that P(x) = 8 − 3x + 2x2+ x4, Q(x) = x5 − 3x4 − 4x2 − 1

and R(x) = 8x3 _{+ 7x}2 _{− 4x then fi nd:}

a P(x) + Q(x) b Q(x) − R(x) c 3P(x) − 2R(x) d 2P(x) − Q(x) + 3R(x). 3 We9 For each of the following polynomials, find: i its degree ii P(0) iii P(2) and iv P(−1).

a P(x) = x6 _{+ 2x}5 _{− x}3 _{+ x}2 _{b P(x) = 3x}7 _{− 2x}6 _{+ x}5 _{− 8} c P(x) = 5x6 _{+ 3x}4 _{− 2x}3 _{− 6x}2 _{+ 3 d P(x) =} −_{7 + 2x − 5x}2 _{+ 2x}3 _{− 3x}4 4 MC If P(x) = x8 − 3x6+ 2x4 − x2+ 3, then P(−2) is equal to:

A 479 B 95 C 31 D 481 E 103 5 We 10 If P(x) = 2x7 + ax5 + 3x3 + bx − 5, P(1) = 4 and P(2) = 163, fi nd a and b.

6 Find a and b, given that f (x) = ax4_{+ bx}3 _{− 3x}2 _{− 4x + 7, f (1) =} −_{2 and f (2) =} −_5. 7 For Q(x) = x5 _{+ 2x}4 _{+ ax}3 _{− 6x + b, Q(2) = 45 and Q(0) =} −_{7. Find a and b.} 8 Find a and b if P(x) = ax6_{+ bx}4_{+ x}3 _{− 6, 3P(1) =} −_{24 and 3P(}−_{2) = 102.} 9 MC

a If P(x) = ax4 _{− x}3 _{+ 3x}2 _{− 5 and P(1) =} −_{1, then a is equal to:}

A 1 B 0 C 2 D −_{3 E} −₂ b If f (x) = xn _{− 2x}3_{+ x}2 _{− 5x and f (2) = 10, then n is equal to:}

A 4 B 6 C 7 D 5 E −₁ exerCISe

1B

On a Calculator page, defi ne the polynomial P(x) by completing the entry line as: Defi ne p(x) = a × x5 _{+ x}4 _{− 3x}3 _{+ b × x − 5} Then press ENTER ·.

To calculate the values of a and b, complete the entry line as:

solve (p(−_{1) =} −_{5 and p(2) =} −_65,a) Then press ENTER ·.

division of polynomials

When one polynomial, P(x), is divided by another, D(x), the result can be expressed as: P x D x Q x R x D x ( ) ( ) ( ) ( ) ( ) = +

where Q(x) is called the quotient, R(x) is called the remainder, and D(x) is called the divisor. Worked exAMple 11

Find the quotient, Q(x), and the remainder, R(x), when x4₋₋₃x3₊₊₂x2_{−− 8} is divided by the linear expression x++ 2.

ThINk WrITe

Method 1: Technology-free

1 Set out the long division with each polynomial in descending powers of x. If one of the powers of x

is missing, include it with 0 as the coeffi cient.

)

x+2 x4 −3x3+2x2+0x−8 2 _{Divide x into x}4 _{and write the result above.}

)

x x x x x

x

+2 4 −3 3+2 2+0 −8

3

3 Multiply the result x3 _{by x + 2 and write the result}

underneath. _x

)

4 _{Subtract and bring down the remaining terms}

to complete the expression. _{x x x x x}

x x x x x + − + + − − + + + − − 2 3 2 0 8 2 5 2 0 8 4 3 2 4 3 3 2 ( )

)

x x x x x x x 3 2 4 3 5 12 24 2 3 2 − + − +

)

division.

7 The polynomial x3 _{− 5x}2_{+ 12x − 24, at the top, is} the quotient.

The quotient, Q(x), is x3 _{− 5x}2_{+ 12x − 24.}

8 _{The result of the fi nal subtraction, 40, is the} remainder.

The remainder, R(x), is 40.

1C

Interactivity int-0246 Division of polynomials eBookplus eBookplus Tutorial int-0517 Worked example 11

Method 2: Technology-enabled 1

2 Write the answer. Dividing x4_{− 3x}3_{+ 2x}2_{− 8 by x + 2 gives a quotient,} Q(x), of x3₋₅x2₊₁₂x_{− and a remainder, R(x), 24} of 40.

Note: P(−2) = (−₂₎4 _{− 3(}−₂₎3 _{+ 2(}−₂₎2 _{− 8} = 16 + 24 + 8 − 8

= 40

The remainder when P(x) is divided by (x + 2) is P(−_2). This leads to the remainder theorem, which states: When P(x) is divided by (x a−− , the remainder is P(a)) or

when P(x) is divided by (ax b_{++ , the remainder is P}) −−

       b a .

Furthermore, if the remainder is zero, then (x − a) is a factor of P(x). This leads to the factor theorem, which states:

If P a( )== 0, then (x a−− is a factor of P(x)) or if (ax b++ is a factor of P(x), then P) −−       == b a 0.

Note: If (x − a) is a factor of P(x) and a is an integer, then a must be a factor of the term independent of x. For example, if (x − 2) is a factor of P(x), then the term independent of x must be divisible by 2. Therefore, (x − 2) could be a factor of x3 _{− 2x}2 _{− x + 2, but (x + 3) could} not be a factor.

Worked exAMple 12

Determine whether or not D x( ) (== x−− 3) is a factor of P x( )==2x3−−4x2−−3x−−8_.

ThINk WrITe

Method 1: Technology-free

1 Evaluate P(3). P (3) = 2(3)3 _{− 4(3)}2 _{− 3(3) − 8}

= 54 − 36 − 9 − 8 = 1

On a Calculator page, press: •  MENu b

•  2:Number 2 •  8:Fraction Tools 8 •  1:Proper Fraction 1 Complete the entry line as: propFrac (x x x ) x 4 ₃ 3 ₂ 2 ₈ 2 − + − + Then press ENTER ·.

2 If P(3) = 0 then (x − 3) is a factor of P(x), but

if P(x) ≠ 0, (x − 3) is not a factor of P(x). P (3) ≠ 0 so (x − 3) is not a factor of P(x). Method 2: Technology-enabled

1

2 Since p(3) = 1, (x − 3) is not a factor of P(x). p( )3 ≠0so(x−3) is not a factor of P(x).

Worked exAMple 13

a Factorise P x( )==2x3−−x2−−13x−−6_{. b Solve 2x}3_−−x2_{−−13x−− ==6 0.}

ThINk WrITe

Method 1: Technology-free

a 1 _{use the factor theorem to find a value for} a where P(a) = 0 and a is a factor of the numerical term. Try a = 1, −_{1, 2,} −_{2, 3,} −_3, 6, −6 until a factor is found.

a P(1) = 2(1)3 _{− (1)}2 _{− 13(1) − 6} = −₁₈ ≠ 0 P(−1) = 2(−₁₎3 _{− (}−₁₎2 _{− 13(}−_{1) − 6} = 4 ≠ 0 P(2) = 2(2)3 _{− (2)}2 _{− 13(2) − 6} = −₂₀ ≠ 0 P(−2) = 2(−₂₎3 _{− (}−₂₎2 _{− 13(}−_{2) − 6} = 0 So (x + 2) is a factor. 2 Divide P(x) by the divisor (x + 2) using long

division.

)

quadratic factors.

P(x) = (x + 2)(2x2 _{− 5x − 3)}

4 Factorise the quadratic, if possible. = (x + 2)(2x + 1)(x − 3) On a Calculator page, define P(x) by

completing the entry lines as: Define p x( )=2x3−4x2− −3x 8 p(3)

b 1 Rewrite the equation in factorised form, using the answer to part a.

b 2x3 _{− x}2 _{− 13x − 6 = 0}

(x + 2)(2x + 1)(x − 3) = 0 2 use the Null Factor Law to state the

solutions.

x = −_2, −1

2 or 3 Method 2: Technology-enabled

a 1 a

2 Write the answer. Factorising p(x) = 2x3_{− x}2_{− 13x − 6 gives} P x( ) (= −x 3)(x+2 2)( x+1)

b 1 b

2 Write the answer. Solving 2x3 _{− x}2 _{− 13x − 6 = 0 gives} x= −₂ x= −1 x= 2 3 , , or P x D x Q x R x D x ( ) ( ) ( ) ( ) ( ) = + 1.

where Q(x) is called the quotient, R(x) is called the remainder, D(x) is called the divisor. Remainder theorem: If

2. P(x) is divided by (x − a), then the remainder is P(a). Factor theorem: If

3. P(a) = 0, then (x − a) is a factor of P(x). If (ax + b) is a factor of P(x), then P b

a

−



 = 0. If (

4. x − a) is a factor of P(x), then a must be a factor of the term independent of x. reMeMBer

On a Calculator page, define the polynomial P(x) by completing the entry line as:

Define p x( )=2x3−x2−13x−6 Then press ENTER ·.

To factorise P(x), complete the entry line as:

factor (p(x))

Then press ENTER ·.

To solve P(x) = 0, complete the entry line as:

solve ( ( )p x = 0, )x Then press ENTER ·.

division of polynomials

1 We 11 Find the quotient, Q(x), and the remainder, R(x), when each of the following polynomials are divided by the given linear expression.

a x3 _{− 2x}2_{+ 5x − 2, x − 4 b x}5 _{− 3x}3_{+ 4x + 3, x + 3} c 6x4 _{− x}3 _{+ 2x}2 _{− 4x, x − 3 d 3x}4 _{− 6x}3 _{+ 12x, 3x + 1} 2 a For each corresponding polynomial in question 1, evaluate:

i P(4) ii P(−₃₎

iii P(3) iv P(− 1₃)

b Compare these values to R(x) in question 1 and comment on the result.

3 We 12 In each of the following determine whether or not D(x) is a factor of P(x). a P(x) = x3 _{+ 9x}2 _{+ 26x − 30, D(x) = x − 3}

b P(x) = x4 _{− x}3 _{− 5x}2 _{− 2x − 8, D(x) = x + 2}

c P(x) = 4 − 9x + 6x2 _{− 13x}3 _{− 12x}4 _{+ 3x}5_{, D(x) = 4 − x} d P(x) = 4x6 _{+ 2x}5 _{− 8x}4 _{− 4x}3 _{+ 6x}2 _{− 9x − 6, D(x) = 2x + 1} 4 MC Examine the equation f (x) = x4 − 4x3 − x2 + 16x − 12.

a Which one of the following is a factor of f (x)?

A x + 1 B x C x + 2 D x + 3 E x − 4 b When factorised, f (x) is equal to:

A (x + 1)(x − 3)(x + 4) B (x + 2)(x − 2)(x − 3)(x − 1)

C (x + 2)(x − 4)(x + 3)(x + 1) D (x − 1)(x + 1)(x − 3)(x − 4)

E x(x − 1)(x + 2)(x + 3)

5 We 13a Factorise the following polynomials.

a P(x) = x3 _{+ 4x}2 _{− 3x − 18 b P(x) = 3x}3 _{− 13x}2 _{− 32x + 12} c P(x) = x4 _{+ 2x}3 _{− 7x}2 _{− 8x + 12 d P(x) = 4x}4 _{+ 12x}3 _{− 24x}2 _{− 32x} 6 We13b Solve each of the following equations.

a 3x3_{+ 3x}2 _{− 18x = 0 b 2x}4_{+ 10x}3 _{− 4x}2 _{− 48x = 0} c 2x4_{+ x}3 _{− 14x}2 _{− 4x + 24 = 0 d x}4 _{− 2x}2_{+ 1 = 0}

7 If (x − 2) is a factor of x3 _{+ ax}2 _{− 6x − 4, then find a.} 8 If (x − 1) is a factor of x3 _{+ x}2 _{− ax + 3, then find a.}

9 Find the value of a if (x + 3) is a factor of 2x4 _{+ ax}3 _{− 3x + 18.}

10 Find the value of a and b if (x + 1) and (x − 2) are factors of ax3 _{− 4x}2 _{+ bx − 12.} 11 If (2x − 3) and (x + 2) are factors of 2x3 _{+ ax}2 _{+ bx + 30, find the values of a and b.}

linear graphs

Linear graphs are polynomials of degree 1. Graphs of linear functions are straight lines and may be sketched by fi nding the intercepts.

revision of properties of straight line graphs

m y y x x = − − 2 1 2 1 exerCISe

1C

1d

2. The general equation of a straight line is:

y = mx + c where m is the gradient

and c is the value of the y-intercept.

3. The equation of a straight line passing through the point (x1, y1) and having a gradient of m is:

y − y1= m(x − x1)

4. The intercept form of the equation of a straight line is: x

a y b

+ = 1 or bx + ay = ab

5. Parallel lines have the same gradient.

6. The product of the gradients of two lines that are perpendicular equals −1. That is, m1 × m2 = −1 or m1 =

−₁ 2

m Worked exAMple 14

Sketch the graph of the linear function 3x−−2y==6 by indicating the intercepts.

ThINk WrITe/drAW

1 Substitute y = 0 into the equation. When y = 0, 3x − 2 × 0 = 6 2 _{Solve the equation for x to find the x-intercept. x = 2} 3 _{Substitute x = 0 into the equation.} Therefore, the x-intercept is 2. 4 Solve the equation for y to find the y-intercept. When x = 0, 3 × 0 − 2y = 6

y = −₃

5 Draw a set of axes. Therefore, the y-intercept is −_3.

6 Indicate the x-intercept and y-intercept and rule a

line through these points. y

x 0 3x − 2y = 6 −3 2 Worked exAMple 15

Find the equation, in the form ax by c++ ++ == 0, of each straight line described below. a The line with a gradient of 2 and passing through ( ,3 2−− )

b The line passing through (0, 8) and ( , )−−_{2 2}

c The line that passes through (3, 4) and is parallel to the line with equation y−−2x−− ==5 0

d The line that passes through (1, 3) and is perpendicular to the line with equation y++2x−− ==3 0 y x 0 A (x₁, y₁) Gradient = m y x 0 (0, b) (a, 0)

ThINk WrITe a 1 Write the rule for the point–gradient

form of the equation of a straight line, y − y1= m(x − x1).

a y − y1= m(x − x1)

2 Substitute the value of the gradient, m, and the coordinates of the point (x1, y1), into the equation.

y − (−_{2) = 2(x − 3)}

3 Expand the brackets. y + 2 = 2x − 6

4 _{Express the equation in the form} required.

y − 2x + 8 = 0 or 2x − y − 8 = 0 b 1 Write the rule for the gradient, m, of

a straight line, given 2 points.

b m y y x x =yy −yy x −x x x 2 1 y₂ y₁ y y 2 1 x₂ x₁ x x

2 Substitute the values of (x1, y1) and (x2, y2) into the rule and evaluate the gradient. = − − − 2 8 2 0 = −₋6 2 = 3 3 Substitute the values of m and (x1, y1)

into the rule for the point–gradient form of the equation of a straight line. (Coordinates of either point given may be used.)

⇒ y − y1= m(x − x1) y − 8 = 3(x − 0)

4 _{Expand the brackets. y − 8 = 3x}

5 Express the equation in the form

required. 3x − y + 8 = 0

c 1 State the gradient of the given line, which is the same as the gradient of the parallel line.

c y − 2x − 5 = 0 becomes y = 2x + 5. The gradient of the parallel lines is 2. 2 Write the rule for the point–gradient

form of the equation of a straight line.

y − y1= m(x − x1)

3 Substitute the values of m and the

coordinates (x₁, y₁) = (3, 4). y − 4 = 2(x − 3) 4 _{Simplify and write in the required form. y − 4 = 2x − 6}

2x − y − 2 = 0 d 1 Find the gradient of the given line. d y = −_{2x + 3}

The gradient of the line is −_2.

2 Find the gradient of the perpendicular line.

The gradient of the perpendicular line is 1 2.

3 Write the rule for the point–gradient form of the equation of a straight line.

y − y1= m(x − x1)

4 _{Substitute the values of m and the} coordinates (x1, y1) = (1, 3).

y − 3 = 1 2(x − 1)

5 Simplify and write in the required form. 2y − 6 = (x − 1) x − 2y + 5 = 0