Ethiopian Grade 10 Mathematics Student Textbook PDF

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Evaluators: Tesfaye Ayele Dagnachew Yalew Tekeste Woldetensai FEDERAL DEMOCRATIC REPUBLIC OF ETHIOPIA MINISTRY OF EDUCATION Published E.C. 2002 by the Federal Democratic Republic of Ethiopia, Ministry of Education, under the General Education Quality Improvement Project (GEQIP) supported by IDA Credit No. 4535-ET, the Fast Track Initiative Catalytic Fund and the Governments of Finland, Italy, Netherlands and the United Kingdom. © 2010 by the Federal Democratic Republic of Ethiopia, Ministry of Education. All rights reserved. No part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means including electronic, mechanical, magnetic or other, without prior written permission of the Ministry of Education or licensing in accordance with the Federal Democratic Republic of Ethiopia, Federal Negarit Gazeta, Proclamation No. 410/2004 – Copyright and Neighbouring Rights Protection. The Ministry of Education wishes to thank the many individuals, groups and other bodies involved – directly and indirectly – in publishing this textbook and the accompanying teacher guide. Copyrighted materials used by permission of their owners. If you are the owner of copyrighted material not cited or improperly cited, please contact with the Ministry of Education, Head Office, Arat Kilo, (PO Box 1367), Addis Ababa, Ethiopia. PHOTO CREDIT: p.1-AA City Government Design and Construction Bureau. p.225 and 271-Encarta Encyclopedia, 2009 edition. p.64-http://www.grupoescolar.com p.172-http://www.ph.surrey.ac.uk Developed and Printed by STAR EDUCATIONAL BOOKS DISTRIBUTORS Pvt. Ltd. 24/4800, Bharat Ram Road, Daryaganj, New Delhi – 110002, INDIA and ASTER NEGA PUBLISHING ENTERPRISE P.O. Box 21073 ADDIS ABABA, ETHIOPIA under GEQIP Contract No. ET-MoE/GEQIP/IDA/ICB/G01/09. ISBN 978-99944-2-044-5  Contents Unit 1 Polynomial Functions............................. 1 1.1 Introduction to polynomial functions....................... 2 1.2 Theorems on polynomials........................................ 19 1.3 Zeros of a polynomial function................................ 27 1.4 Graphs of polynomial functions.............................. 36 Key Terms................................................... 50 Summary.................................................... 50 Review Exercises on Unit 1........................... 51 Unit 2 Exponential and Logarithmic Functions. 53 5 4 y y=e y = ex x y=x y=x 2.1 Exponents and logarithms....................................... 54 3 2.2 The exponential functions and their graphs........... 75 2 y = ln x y = lnx 1 x 2.3 The logarithmic functions and their graphs............ 84 -4 -3 -2 -1 -1 1 2 3 4 5 2.4 Equations involving exponents and -2 logarithms................................................................ 92 -3 -4 2.5 Applications of exponential and logarithmic functions.................................................................... 98 Key Terms.................................................. 104 Summary................................................... 104 Review Exercises on Unit 2.......................... 106 Unit 3 Solving Inequalities.............................. 111 y 4 3.1 Inequalities involving absolute value.................... 112 3 2 3.2 Systems of liner inequalities in two variables........ 119 1 3.3 Quadratic inequalities............................................. 127 x 3 -2 -1 1 2 3 4 Key Terms.................................................. 140 -1 -2 Summary................................................... 140 Review Exercises on Unit 3.......................... 141 Unit 4 Coordinate Geometry........................... 143 4.1 Distance between two points................................. 144 4.2 Division of a line segment....................................... 148 4.3 Equation of a line..................................................... 153 4.4 Parallel and perpendicular lines............................ 165  Key Terms...................................................... 169 Summary....................................................... 169 Review Exercises on Unit 4.............................. 170 Unit 5 Trigonometric Functions...................... 171 f(x)= f(x) sin x  = Sin(x) 5.1 Basic trigonometric functions................................. 172 y f(x)  ==cos(x) cos x 1 f(x) 5.2 The reciprocal functions of the basic 0.5 trigonometric functions.......................................... 201 0 x - 0.5 5.3 Simple trigonometric identities............................... 207 -1 -2π π - 0 π π 2π 5.4 Real life application problems............................... 213 Key Terms...................................................... 219 Summary....................................................... 219 Review Exercises on Unit 5.............................. 222 Unit 6 Plane Geometry................................... 225 6.1 Theorems on triangles............................................. 226 6.2 Special quadrilaterals............................................. 239 6.3 More on circles........................................................ 246 6.4 Regular polygons.................................................... 261 Key Terms...................................................... 265 Summary....................................................... 266 Review Exercises on Unit 6.............................. 267 Unit 7 Measurement....................................... 269 7.1 Revision on surface areas and volumes of prisms and cylinders........................................................... 270 7.2 Pyramids, cones and spheres................................ 277 7.3 Frustums of pyramids and cones........................... 284 7.4 Surface areas and volumes of composed solids.. 294 Key Terms...................................................... 298 Summary....................................................... 298 Review Exercises on Unit 7.............................. 299 Table of Trigonometric Functions..................................... 301 Table of Common Logarithms......................................... 302 Unit Polynomial functions are the most widely used functions in Mathematics. They arise naturally in many applications. Essentially, the graph of a polynomial function has no breaks and gaps. It describes smooth curves as shown in the figure above. POLYNOMIAL FUNCTIONS Unit Outcomes: After completing this unit, you should be able to: define polynomial functions. perform the four fundamental operations on polynomials. apply theorems on polynomials to solve related problems. determine the number of rational and irrational zeros of a polynomial. sketch and analyse the graphs of polynomial functions. Main Contents 1.1 Introduction to polynomial functions 1.2 Theorems on polynomials 1.3 Zeros of polynomial functions 1.4 Graphs of polynomial functions Key Terms Summary Review Exercises Mathematics Grade 10 INTRODUCTION There is an extremely important family of functions in mathematics called polynomial functions. Stated quite simply, polynomial functions are functions with x as an input variable, consisting of the sum of several terms, each term is a product of two factors, the first being a real number coefficient and the second being x raised to some non-negative integer power. In this unit you will be looking at the different components of polynomial functions. These are theorems on polynomial functions; zeros of a polynomial function; and graphs of polynomial functions. Basically the graph of a polynomial function is a smooth and continuous curve. However, you will be going over how to use its degree (even or odd) and the leading coefficient to determine the end behaviour of its graph. 1.1 INTRODUCTION TO POLYNOMIAL FUNCTIONS
OPENING PROBLEM Obviously, the volume of water in any dam fluctuates from season to season. An engineer suggests that the volume of the water (in giga litres) in a certain dam after t-months (starting 1st September) is described by the model: v (t) = 450 – 170t + 22t2 – 0.6t3 Electric Power Corporation rules that if the volume falls below 200 giga litres, its side- wise project, “irrigation”, is prohibited. During which months, if any, was irrigation prohibited in the last 12 months? Recall that, a function f is a relation in which no two ordered pairs have the same first element, which means that for any given x in the domain of f, there is a unique pair (x, y) belonging to the function f. In Unit 4 of Grade 9 mathematics, you have discussed functions such as: 2 1 f ( x) = x + , g (x) = 5 – 3x, h (x) = 8x and l ( x) = − 3x + 2.7. 3 2 Such functions are linear functions. A function f is a linear function, if it can be written in the form f (x) = ax + b, a ≠ 0, where a and b are real numbers. 2  Unit 1 Polynomial Functions The domain of f is the set of all real numbers and the range is also the set of all real numbers. If a = 0, then f is called a constant function. In this case, f (x) = b. This function has the set of all real numbers as its domain and {b} as its range. Also recall what you studied about quadratic functions. Each of the following functions is a quadratic function. 1 2 f (x) = x2 + 7x – 12, g (x) = 9 + x , h (x) = – x2 + π, k (x) = x2, 4 l (x) = 2 (x – 1)2 + 3, m (x) = (x + 2) (1 – x) If a, b, and c are real numbers with a ≠ 0, then the function f (x) = ax2 + bx + c is a quadratic function. Since the expression ax2 + bx + c represents a real number for any real number x, the domain of a quadratic function is the set of all real numbers. The range of a quadratic function depends on the values of a , b and c. Exercise 1.1 1 In each of the following cases, classify the function as constant, linear, quadratic, or none of these: a f (x) = 1 – x2 b h (x) = 2 x−1 4 c h (x) = 3 + 2x d g (x) = 5 – x 5 −1 2 e f (x)= 2 3 f f (x) =   3 g h (x) = 1 − | x | h f (x) = (1 – 2 x ) (1 + 2 x) 3 i k (x) = (12 + 8x) j f (x) = 12x−1 4 ( x + 1) (x − 2) k l ( x) = l f ( x) = x4 − x +1 x − 2 2 For what values of a, b, and c is f (x) = ax2 + bx + c a constant, a linear or a quadratic function? 3 Mathematics Grade 10 1.1.1 Definition of a Polynomial Function Constant, linear and quadratic functions are all special cases of a wider class of functions called polynomial functions. Definition 1.1 Let n be a non-negative integer and let an, an – 1,..., a1, ao be real numbers with an ≠ 0. The function p(x) = anxn + an – 1 xn – 1 +... + a1 x + ao is called a polynomial function in variable x of degree n. Note that in the definition of a polynomial function p (x) = anxn + an – 1xn – 1 +... + a1x + ao i an , an – 1, an – 2,..., a1 , ao are called the coefficients of the polynomial function (or simply the polynomial). ii The number an is called the leading coefficient of the polynomial function and anxn is the leading term. iii The number a0 is called the constant term of the polynomial. iv The number n (the exponent of the highest power of x), is the degree of the polynomial. Note that the domain of a polynomial function is R. Example 1 Which of the following are polynomial functions? For those which are polynomials, find the degree, leading coefficient, and constant term. 2 7 x a f (x) = x4 – 12x2 + x + b f (x) = 3 8 x c g (x) = ( x + 1) 2 d f (x) = 2x–4 + x2 + 8x + 1 x2 + 1 8 15 e k (x) = f g (x) = x x2 + 1 5 6 g f (x) = (1 – 2 x ) (1 + 2 x) h k ( y) = y Solution: 2 a It is a polynomial function of degree 4 with leading coefficient and 3 7 constant term. 8 b It is not a polynomial function because its domain is not R. 4  Unit 1 Polynomial Functions 2 c g ( x) = ( x + 1) = x + 1 , so it is not a polynomial function because it cannot be written in the form g (x) = anxn + an – 1 xn – 1 +... + a1 x + ao. d It is not a polynomial function because one of its terms has a negative exponent. x2 + 1 e k (x) = 2 = 1, so it is a polynomial function of degree 0 with leading x +1 coefficient 1 and constant term 1. 8 f It is a polynomial function of degree 15 with leading coefficient and 5 constant term 0. g It is a polynomial function of degree 2 with leading coefficient – 2 and constant term 1. h It is not a polynomial function because its domain is not R. A polynomial expression in x is an expression of the form an x n + an −1 x n −1 +... + a1 x + ao where n is a non negative integer and an≠ 0. Each individual expression ak xk making up the polynomial is called a term. ACTIVITY 1.1 x2 + 3 − 6 x4 7 1 For the polynomial expression + x − x3 , 4 8 a what is the degree? b what is the leading coefficient? c what is the coefficient of x3? d what is the constant term? 2 A match box has length x cm, width x + 1 cm and height 3 cm, a Express its surface area as a function of x. b What is the degree and the constant term of the polynomial obtained above? We can restate the definitions of linear and quadratic functions using the terminology for polynomials. Linear functions are polynomial functions of degree 1. Nonzero constant functions are polynomial functions of degree 0. Similarly, quadratic functions are polynomial functions of degree 2. The zero function, p (x) = 0, is also considered to be a polynomial function but is not assigned a degree at this level. Note that in expressing a polynomial, we usually omit all terms which appear with zero coefficients and write others in decreasing order, or increasing order, of their exponents. 5 Mathematics Grade 10 x 2 − 2 x5 + 8 7 Example 2 For the polynomial function p(x) = + x − x3 , 4 8 a what is its degree? b find an, an – 1 , an – 2 and a2. c what is the constant term? d what is the coefficient of x? x 2 − 2 x5 + 8 7 3 x2 2 5 8 7 Solution: p(x) = + x−x = − x + + x − x3 4 8 4 4 4 8 1 5 3 1 2 7 = – x −x + x + x +2 2 4 8 a The degree is 5. −1 1 b an = a5 = , an –1 = a4 = 0, an – 2 = a3 = –1 and a2 =. 2 4 c The constant term is 2. 7 d The coefficient of x is. 8 Although the domain of a polynomial function is the set of all real numbers, you may have to set a restriction on the domain because of other circumstances. For example, in a geometrical application, if a rectangle is x centimetres long, and p (x) is the area of the rectangle, the domain of the function p is the set of positive real numbers. Similarly, in a population function, the domain is the set of positive integers. has, a polynomial function p is said to Based on the types of coefficients it has, be:  a polynomial function over integers, if the coefficients of p (x) are all integers.  a polynomial function over rational numbers, if the coefficients of p (x) are all rational numbers.  a polynomial function over real numbers, if the coefficients of p (x) are all real numbers. Remark: Every polynomial function that we will consider in this unit is a polynomial function over the real numbers. 2 4 7 For example, if g ( x ) =x − 13 x 2 + , then g is a polynomial function over rational and 3 8 real numbers, but not over integers. If p (x) can be written in the form, anxn + an –1 xn – 1 +... + a1 x + ao, then different expressions can define the same polynomial function. 6  Unit 1 Polynomial Functions For example, the following expressions all define the same polynomial function as 1 2 x −x. 2 x2 − 2 x 1 1 2 1  a b − x + x2 c ( x − 2 x) d x  x − 1 2 2 2 2  Any expression which defines a polynomial function is called a polynomial expression. Example 3 For the polynomial expression 6x3 – x5 + 2x + 1, a what is the degree? b what is the coefficient of x3? c what is the leading coefficient? d what is the constant term? Solution: a The degree is 5. b The coefficient of x3 is 6. c The leading coefficient is – 1. d The constant term is 1. Consider the functions f ( x) = ( x + 3)( x − 1) and g(x) = x + 3. x −1 When f is simplified it gives f (x) = x + 3, where x ≠ 1. As the domain of f is not the set of all real numbers, f is not a polynomial function. But the domain of g is the set of all real numbers. The functions f and g have different domains and you can conclude that f and g are not the same functions. When you are testing an expression to check whether or not it defines a polynomial function, you must be careful and watch the domain of the function defined by it. Exercise 1.2 1 Which of the following are polynomial functions? a f (x) = 3x4 – 2x3 + x2 + 7x – 9 b f (x) = x25 + 1 1 2 2 c f (x) = 3x–3 + 2x–2 + x + 4 d f (y) = y + y+1 3 3 3 2 e f (t ) = + 2 f f (y) = 108 – 95y t t g f (x) = 312x6 h f (x) = 3x2 − x3 + 2 7 Mathematics Grade 10 4 x 2 − 5 x3 + 6 i f (x) = 3x + x + 3 j f (x) = 8 3 18 k f (x) = l f (y) = 6+ x y a x m f (a) = n f (x) = 2a 12 1 o f (x) = 0 p f (a) = a 2 + 3a + a2 9 83 4 q f (x) = x + 54 x97 + π r f (t) = − 2π 17 7  2  3 s f (x) = (1 – x) (x + 2) t g ( x) =  x −  x +   3  4 2 Give the degree, the leading coefficient and the constant term of each polynomial function in Question 1 above. 3 Which of the polynomial functions in Question 1 above are: a polynomial functions over integers? b polynomial functions over rational numbers? c polynomial functions over real numbers? 4 Which of the following are polynomial expressions? ( x + 3) 2 a 2 3−x b y (y – 2) c x+3 ( y − 3) ( y −1) (t − 5) (t − 1) d y 2 + 3 + 2 − 3y3 e f 2 t −1 ( x − 3) ( x 2 + 1) x2 + 4 g h y + 2y – 3y i x2 + 1 x2 + 4 5 An open box is to be made from a 20 cm long square piece of material, by cutting equal squares of length x cm from the corners and turning up the sides as shown in Figure 1.1. x x a Verify that the volume of the box is given x by the function v (x) = 4x3 – 80x2 + 400x. b Determine the domain of v. Figure 1.1 8  Unit 1 Polynomial Functions 1.1.2 Operations on Polynomial Functions Recall that, in algebra, the fundamental operations are addition, subtraction, multiplication and division. The first step in performing operations on polynomial functions is to use the commutative, associative and distributive laws in order to combine like terms together. ACTIVITY 1.2 1 What are like terms? Give an example. 2 Are 8a2, 2a3 and 5a like terms? Explain. 3 For any three real numbers a, b and c, determine whether each of the following statements is true or false. Give reasons for your answers. a a – (b + c) = a – b + c b a + (b – c) = a + b – c c a – (b – c) = a – b + c d a – (b – c ) = a – b – c 4 Verify each of the following statements: a (4x + a) + (2a – x) = 3 (a + x) b 5x2 y + 2xy2 – (x2y – xy2) = 4x2y + 3xy2 c 8a – (b + 9a) = – (a + b) d 2x – 4 (x – y) + (y – x) = 5y – 3x 5 If f (x) = x3 – 2x2 + 1 and g (x) = x2 – x – 1, then which of the following statements are true? a f (x) + g (x) = x3 + x2 – x b f (x) – g (x) = x3 – 3x2 + x + 2 c g (x) – f (x) = 3x2 + x3 – x – 2 d f (x) – g (x) ≠ g (x) – f (x). 6 If f and g are polynomial functions of degree 3, then which of the following is necessarily true? a f + g is of degree 3. b f + g is of degree 6. c 2f is of degree 3. d fg is of degree 6. Addition of polynomial functions You can add polynomial functions in the same way as you add real numbers. Simply add the like terms by adding their coefficients. Note that like terms are terms having the same variables to the same powers but possibly different coefficients. 9 Mathematics Grade 10 For example, if f (x) = 5x4 – x3 + 8x – 2 and g (x) = 4x3 – x2 − 3x + 5, then the sum of f (x) and g (x) is the polynomial function: f (x) + g (x) = (5x4 – x3 + 8x – 2) + (4x3 – x2 – 3x + 5) = 5x4 + (–x3 + 4x3) – x2 + (8x – 3x) + (–2 + 5)... (grouping like terms) = 5x4 + (4 – 1)x3 – x2 + (8 – 3) x + (5 – 2)...... (adding their coefficients) = 5x4 + 3x3 – x2 + 5x + 3................... (combining like terms). Therefore, the sum f (x) + g (x) = 5x4 + 3x3 – x2 + 5x + 3 is a polynomial of degree 4. The sum of two polynomial functions f and g is written as f + g, and is defined as: f + g : (f + g) (x) = f (x) + g (x), for all x ∈ ℝ. Example 4 In each of the following, find the sum of f (x) and g (x): 2 2 1 1 a f (x) = x3 + x – x + 3 and g (x) = −x3 + x2 + x − 4. 3 2 3 b f (x) = 2x5 + 3x4 – 2 2x3 + x – 5 and g (x) = x4 + 2x3 + x2 + 6x + 8. Solution: 2 2 1  3 1 2  a f (x) + g (x) = (x3 + x − x + 3) +  −x + x + x − 4  3 2  3  2 1   1  = (x3 – x3) +  x 2 + x 2  +  − x + x  + (3 − 4)... (grouping like terms) 3 3   2   2 1  1 = (1 – 1)x3 +  +  x 2 + 1 −  x + ( 3 − 4 ).. (adding their coefficients)  3 3  2 1 = x2 + x − 1.......................... (combining like terms) 2 1 So, f (x) + g (x) = x2 + x – 1, which is a polynomial of degree 2. 2 b f (x) + g (x) = (2x5 + 3x4 – 2 2x3 + x – 5) + (x4 + 2 x3 + x2 + 6 x + 8) = 2x5 + (3x4 + x4) + (_2 2 x3 + 2 x3 ) + x 2 + (x + 6x) + ( − 5 + 8) = 2x5 + (3 + 1)x4 + (–2 2 + 2 )x3 + x2 + (1 + 6) x + (8 – 5) = 2x5 + 4x4 – 2 x3 + x2 + 7x + 3 So, f (x) + g (x) = 2x5 + 4x4 – 2 x3 + x2 + 7x + 3, which is a polynomial function of degree 5. 10  Unit 1 Polynomial Functions ACTIVITY 1.3 1 What do you observe in Example 4 about the degree of f + g? 2 Is the degree of (f + g) (x) equal to the degree of f (x) or g (x), whichever has the highest degree? 3 If f (x) and g (x) have same degree, then the degree of (f + g) (x) might be lower than the degree of f (x) or the degree of g (x).Which part of Example 4 illustrates this situation? Why does this happen? 4 What is the domain of (f + g) (x)? Subtraction of polynomial functions To subtract a polynomial from a polynomial, subtract the coefficients of the corresponding like terms. So, whichever term is to be subtracted, its sign is changed and then the terms are added. For example, if f (x) = 2x3 – 5x2 + x – 7 and g (x) = 8x2 – x3 + 4x + 5, then the difference of f (x) and g (x) is the polynomial function: f (x) – g (x) = (2x3 – 5x2 + x – 7) – (8x2 – x3 + 4x + 5) = 2x3 – 5x2 + x – 7 – 8x2 + x3 – 4x – 5...... (removing brackets) = (2 + 1) x3+(–5 – 8) x2+(1 – 4)x + (–7 – 5)..(adding coefficients of like terms) = 3x3 – 13x2 – 3x – 12..................(combining like terms) The difference of two polynomial functions f and g is written as f – g, and is defined as: (f − g) : (f – g) (x) = f (x) − g (x), for all x ∈ ℝ. Example 5 In each of the following, find f – g; a f (x) = x4 + 3x3 – x2 + 4 and g (x) = x4 – x3 + 5x2 + 6x b f (x) = x5 + 2x3 – 8x + 1 and g (x) = x3 + 2x2 + 6x – 9 Solution: a f (x) – g (x) = (x4 + 3x3 – x2 + 4) – (x4 – x3 + 5x2 + 6x) = x4 + 3x3 – x2 + 4 –x4 + x3 –5x2 – 6x……....(removing brackets) = (1 – 1)x4 + (3 + 1)x3 + (–1 – 5)x2 – 6x + 4..(adding their coefficients) = 4x3 – 6x2 – 6x + 4......................................(combining like terms) Therefore, the difference is a polynomial function of degree 3, f (x) – g (x) = 4x3 – 6x2 – 6x + 4 11 Mathematics Grade 10 b f (x) – g (x) = (x5 + 2x3 – 8x + 1) – (x3 + 2x2 + 6x – 9 ) = x5 + 2x3 – 8x + 1 – x3 – 2x2 – 6x + 9 = x5 + (2x3 – x3) – 2x2 + (–8x – 6x) + (1 + 9) = x5 + (2 – 1)x3 – 2x2 + (–8 – 6) x + (1 + 9) = x5 + x3 – 2x2 – 14x + 10 Therefore the difference f (x) – g (x) = x5 + x3 – 2x2 – 14x + 10, which is a polynomial function of degree 5. Note that if the degree of f is not equal to the degree of g, then the degree of (f – g) (x) is the degree of f (x) or the degree of g (x), whichever has the highest degree. If they have the same degree, however, the degree of (f – g) (x) might be lower than this common degree when they have the same leading coefficient as illustrated in Example 5a. Multiplication of polynomial functions To multiply two polynomial functions, multiply each term of one by each term of the other, and collect like terms. For example, let f (x) = 2x3 – x2 + 3x – 2 and g (x) = x2 – 2x + 3. Then the product of f (x) and g (x) is a polynomial function: f (x). g (x) = (2x3 – x2 + 3x – 2)⋅(x2 – 2x + 3) = 2x3(x2 − 2x + 3) − x2(x2 − 2x + 3) + 3x(x2−2x + 3)−2(x2 − 2x +3) = 2x5 – 4x4 + 6x3 – x4 +2x3 – 3x2 +3x3 – 6x2 + 9x – 2x2 + 4x – 6 = 2x5 + (–4x4 – x4) + (6x3 + 2x3 + 3x3) + (–3x2 – 6x2 – 2x2) + (9x + 4x) – 6 = 2x5 – 5x4 + 11x3 – 11x2 + 13x – 6 The product of two polynomial functions f and g is written as f⋅g, and is defined as: f⋅g : (f⋅g) (x) = f (x)⋅g(x), for all x ∈ ℝ. Example 6 In each of the following, find f. g and give the degree of f. g : 3 2 9 a f (x) = x + , g (x) = 4xb f (x) = x2 + 2x, g (x) = x5 + 4x2 – 2 4 2 3 9 Solution: a f (x).g (x) =  x 2 + .(4 x) = 3x3+18x 4 2 So, the product (f.g) (x) = 3x3 + 18x has degree 3. 12  Unit 1 Polynomial Functions b f (x).g (x) = (x2 + 2x).(x5 + 4x2 – 2) = x2 (x5 + 4x2 – 2) + 2x (x5 + 4x2 – 2) = x7 + 2x6 + 4x4 + 8x3 – 2x2 – 4x So, the product (f⋅g) (x) = x7 + 2x6 + 4x4 + 8x3 – 2x2 – 4x has degree 7. In Example 6, you can see that the degree of f⋅g is the sum of the degrees of the two polynomial functions f and g. To find the product of two polynomial functions, we can also use a vertical arrangement for multiplication. Example 7 Let f (x) = 3x2 – 2x3 + x5 – 8x + 1 and g (x) = 5 + 2x2 + 8x. Find f (x). g (x) and the degree of the product. Solution: To find the product, f.g, first rearrange each polynomial in descending powers of x as follows: x5 – 2x3 + 3x2 – 8x + 1 Like terms are written 2x2 + 8x + 5 in the same column. 5x5 + 0x4 –10x3 + 15x2 – 40x + 5.....(multiplying by 5) 8x6 + 0x5 – 16x4 + 24x3 – 64x2 + 8x........(multiplying by 8x) 2x7 + 0x6 – 4x5 + 6x4 – 16x3 + 2x2 …. (multiplying by 2x2) 2x7 + 8x6 + x5 – 10x4 – 2x3 – 47x2 – 32x + 5.......(adding vertically.) Thus, f (x).g (x) = 2x7 + 8x6 + x5 – 10x4 – 2x3 – 47x2 – 32x + 5 and hence the degree of f.g is 7. ACTIVITY 1.4 1 For any non-zero polynomial function, if the degree of f is m and the degree of g is n, then what is the degree of f.g? 2 If either f or g is the zero polynomial, what is the degree of f.g? 3 Is the product of two or more polynomials always a polynomial? Example 8 (Application of polynomial functions) A person wants to make an open box by cutting equal squares from the corners of a piece of metal 160 cm by 240 cm as shown in Figure 1.2. If the edge of each cut-out square is x cm, find the volume of the box, when x = 1 and x = 3. 13 Mathematics Grade 10 160 cm 160 – 2x x 240 – 2x x x x 240 cm x 160 – 2x 240 – 2x Figure 1.2 Solution: The volume of a rectangular box is equal to the product of its length, width and height. From the Figure 1.2, the length is 240 – 2x, the width is 160 – 2x, and the height is x. So the volume of the box is v (x) = (240 – 2x) (160 – 2x) (x) = (38400 – 800x + 4x2) (x) = 38400x – 800x2 + 4x3(a polynomial of degree 3) When x = 1, the volume of the box is v (1) = 38400 – 800 + 4 = 37604 cm3 When x = 3, the volume of the box is v (3) = 38400 (3) – 800 (3)2 + 4 (3)3 = 115200 – 7200 + 108 = 108,108 cm3 Division of polynomial functions It is possible to divide a polynomial by a polynomial using a long division process similar to that used in arithmetic. Look at the calculations below, where 939 is being divided by 12. 7 78 12 939 12 939 84 84 99 99 96 3 14  Unit 1 Polynomial Functions The second division can be expressed by an equation which says nothing about division. 939 3 939 = (78 × 12) + 3. Observe that, 939 ÷ 12 = 78 + (3÷12) or = 78 +. 12 12 Here 939 is the dividend, 12 is the divisor, 78 is the quotient and 3 is the remainder of the division. What we actually did in the above calculation was to continue the process as long as the quotient and the remainder are integers and the remainder is less than the divisor. ACTIVITY 1.5 x2 − x + 2 4 1 Consider the following: f (x) = = x +1+. Which x−2 x−2 polynomials do you think we should call the divisor, dividend, quotient and remainder? 2 Divide x3 + 1 by x + 1. (You should see that the remainder is 0) 3 When do we say the division is exact? 4 What must be true about the degrees of the dividend and the divisor before you can try to divide polynomials? 5 Suppose the degree of the dividend is n and the degree of the divisor is m. If n > m, then what will be the degree of the quotient? When should we stop dividing one polynomial by another? Look at the three calculations below: 3 x x+2 x+2+ x +1 x+1 x2 + 3x + 5 x + 1 x2 + 3x + 5 2 x + 1 x + 3x + 5 x2 + x x2 + x x2 + x 2x + 5 2x + 5 2x + 5 2x + 2 2x + 2 3 3 3 0 The first division above tells us that x2 + 3x + 5 = x (x + 1) + 2x + 5. It holds true for all values of x ≠ –1. In the middle one of the three divisions, you continued as long as you got a quotient and remainder which are both polynomials. 15 Mathematics Grade 10 When you are asked to divide one polynomial by another, stop the division process when you get a quotient and remainder that are polynomials and the degree of the remainder is less than the degree of the divisor. Study the example below to divide 2x3 – 3x2 + 4x + 7 by x – 2. x2 Think =x x 2x 3 6x Think = 2x2 Think =6 x x 2x2 + x + 6 Quotient Divisor x – 2 2x3 – 3x2 + 4x + 7 Dividend 2x3 – 4x2 multiply 2x2 (x – 2) x2 + 4x + 7 subtract x2 – 2x multiply x (x – 2) 6x + 7 subtract 6x – 12 multiply 6 (x – 2) Remainder 19 subtract So, dividing 2x3 – 3x2 + 4x + 7 by x – 2 gives a quotient of 2x2 + x + 6 and a 2 x3 − 3x2 + 4 x + 7 19 remainder of 19. That is, = 2x2 + x + 6 + x−2 x−2 The quotient (division) of two polynomial functions f and g is written as f ÷ g, and is defined as: f ÷ g : ( f ÷ g) (x) = f (x) ÷ g (x), provided that g (x) ≠ 0, for all x ∈ ℝ. Example 9 Divide 4x3 – 3x + 5 by 2x – 3 Solution: 2x2 + 3x + 3 Arrange the dividend and the divisor in 3 2 descending powers of x. 2x – 3 4x + 0x – 3x + 5 Insert (with 0 coefficients) for missing terms. 4x3 – 6x2 Divide the first term of the dividend by the first 6x2 – 3x + 5 term of the divisor. 6x2 – 9x Multiply the divisor by 2x2, line up like terms 6x + 5 and, subtract Repeat the process until the degree of the 6x – 9 remainder is less than that of the divisor. Remainder 14 Therefore, 4x3 – 3x + 5 = (2x2 + 3x + 3) (2x – 3) + 14 16  Unit 1 Polynomial Functions Example 10 Find the quotient and remainder when x5 + 4x3 – 6x2 – 8 is divided by x2 + 3x + 2. Solution: x 3 − 3 x 2 + 11x − 33 x 2 + 3 x + 2 x5 + 0 x 4 + 4 x3 − 6 x 2 + 0 x − 8 x5 + 3x 4 + 2 x3 −3 x 4 + 2 x 3 − 6 x 2 + 0 x − 8 −3 x 4 − 9 x 3 − 6 x 2 11x 3 + 0 x 2 + 0 x − 8 11x 3 + 33 x 2 + 22 x −33 x 2 − 22 x − 8 −33 x 2 − 99 x − 66 77 x + 58 3 2 Therefore the quotient is x – 3x + 11x – 33 and the remainder is 77x + 58 x5 + 4 x3 − 6 x 2 − 8 77 x + 58 We can write the result as 2 = x3 − 3x2 + 11x − 33 + 2. x + 3x + 2 x + 3x + 2 Group Work 1.1 1 Find two polynomial functions f and g both of degree three with f + g of degree one. What relations do you observe between the leading coefficients of f and g? f 2 Given f (x) = x + 2 and g (x) = ax + b, find all values of a and b so that is a g polynomial function. 2 2 3 Given polynomial functions g (x) = x + 3, q (x) = x − 5 and r ( x) = 2 x + 1 , find a f ( x) r ( x) function f (x) such that = q ( x) +. g ( x) g ( x) Exercise 1.3 1 Write each of the following expressions, if possible, as a polynomial in the form anxn + an – 1 xn – 1 +... + a1x + ao: a (x2 – x – 6) – (x + 2) b (x2 – x – 6)(x + 2) 2 x2 − x − 6 c (x + 2) – (x – x – 6) d x+2 x+2 e 2 f (x2 – x – 6)2 x − x−6 x–3 g 2 + 23 – x h (2x + 3)2 i (x2 – x + 1)(x2 – 3x + 5) j (x3 – x4 + 2x + 1) – (x4 + x3 – 2x2 + 8) 17 Mathematics Grade 10 2 Let f and g be polynomial functions such that f (x) = x2 – 5x + 6 and g (x) = x2 – x + 3. Which of the following functions are also polynomial functions? f a f+g b g–f c f⋅g d g 2 e f −g f 2f + 3g g f2 3 If f and g are any two polynomial functions, which of the following will always be a polynomial function? f a f+g b f–g c f⋅g d g 3 1 f −g e f2 f g− f g 4 3 f +g 4 In each of the following, find f + g and f – g and give the degree of f, the degree of g, the degree of f + g and the degree of f – g: 2 a f (x) = 3x – ; g (x) = 2x + 5 3 b f (x) = –7x2 + x – 8; g (x) = 2x2 – x + 1 c f (x) = 1 – x3 + 6x2 – 8x; g (x) = x3 + 10 5 In each of the following, i find the function f⋅g. ii give the degree of f and the degree of g. iii give the degree of f⋅g. a f (x) = 2x + 1; g (x) = 3x – 5 b f (x) = x2 – 3x + 5; g (x) = 5x + 3 c f (x) = 2x3 – x – 7; g (x) = x2 + 2x d f (x) = 0; g (x) = x3 – 8x2 + 9 6 In each of the following, divide the first polynomial by the second: a x3 – 1; x – 1 b x3 + 1; x2 – x + 1 c x4 – 1; x2 + 1 d x5 + 1; x + 1 e 2x5 – x6 + 2x3 + 6; x3 – x – 2 7 For each of the following, find the quotient and the remainder: a (5 – 6x + 8x2) ÷ (x – 1) b (x3 – 1) ÷ (x – 1) c (3y – y2 + 2y3 – 1) ÷ (y2 + 1) d (3x4 + 2x3 – 4x – 1) ÷ (x + 3)  2 e (3x3 – x2 + x + 2) ÷  x +   3 18  Unit 1 Polynomial Functions 1.2 THEOREMS ON POLYNOMIALS 1.2.1 Polynomial Division Theorem Recall that, when we divided one polynomial by another, we apply the long division procedure, until the remainder was either the zero polynomial or a polynomial of lower degree than the divisor. For example, if we divide x2 + 3x + 7 by x + 1, we obtain the following. x+2 quotient 2 Divisor x + 1 x + 3x + 7 dividend 2 x +x 2x + 7 2x + 2 5 remainder In fractional form, we can write this result as follows: dividend quotient remainder x 2 + 3x + 7 5 =x+2+ x +1 x +1 divisor divisor This implies that x2 + 3x + 7 = (x + 1) (x + 2) + 5 which illustrates the theorem called the polynomial division theorem. ACTIVITY 1.6 1 For each of the following pairs of polynomials, find q (x) and r (x) that satisfy f (x) = d (x) q (x) + r (x). a f (x) = x2 + x – 7; d (x) = x – 3 b f (x) = x3 – x2 + 8; d (x) = x + 2 c f (x) = x4 – x3 + x – 1; d (x) = x – 1 2 In Question 1, what did you observe about the degrees of the polynomial functions f (x) and d (x)? f ( x) 3 In Question 1, the fractional expression is improper. Why? d ( x) r ( x) 4 Is proper or improper? What can you say about the degree of r (x) and d (x)? d ( x) 19 Mathematics Grade 10 Theorem 1.1 Polynomial division theorem If f(x) and d(x) are polynomials such that d(x) ≠ 0, and the degree of d(x) is less than or equal to the degree of f(x), then there exist unique polynomials q(x) and r(x) such that f(x) = d(x) q(x) + r(x) Dividend Quotient Divisor Remainder where r(x) = 0 or the degree of r (x) is less than the degree of d(x). If the remainder r(x) is zero, f(x) divides exactly into d(x). Proof:- i Existence of the polynomials q (x) and r (x ) Since f (x) and d (x) are polynomials, long division of f (x) by d (x) will give a quotient q(x) and remainder r (x), with degree of r (x) < degree of d (x) or r (x) = 0. ii The uniqueness of q (x) and r (x ) To show the uniqueness of q (x) and r (x), suppose that f (x) = d (x)q1(x) + r1(x) and also f (x) = d (x)q2(x) + r2(x) with deg r1(x) < deg d (x) and deg r2(x) < deg d(x). Then r2(x) = f (x) – d (x) q2(x) and r1(x) = f (x) – d (x) q1(x) ⇒ r2(x) – r1(x) = d (x) [q1(x) – q2(x)] Therefore, d (x) is a factor of r2(x) – r1(x) As deg (r2(x) – r1(x)) ≤ max {deg r1 ( x), deg r2 ( x)} < deg d (x) it follows that, r2(x) – r1(x) = 0 As a result r1(x) = r2(x) and q1(x) = q2(x). Therefore, q (x) and r (x) are unique polynomial functions. 20  Unit 1 Polynomial Functions Example 1 In each of the following pairs of polynomials, find polynomials q (x) and r (x) such that f (x) = d (x) q (x) + r (x). a f (x) = 2x3 – 3x + 1; d (x) = x + 2 b f (x) = x3 – 2x2 + x + 5; d (x) = x2 + 1 c f (x) = x4 + x2 – 2 ; d (x) = x2 – x + 3 Solution: f ( x) 2 x 3 − 3x + 1 9 a = = 2 x2 − 4x + 5 − d ( x) x+2 x+2 ⇒ 2x3 − 3x + 1 = (x+2)(2x2 − 4x + 5) − 9 Therefore q (x) = 2x2 – 4x + 5 and r (x) = − 9. f ( x) x3 − 2 x 2 + x + 5 7 b = 2 = x−2+ 2 d ( x) x +1 x +1 ⇒ x3 – 2x2 + x + 5 = (x2 + 1) (x – 2) + 7 Therefore q (x) = x – 2 and r (x) = 7. f ( x) x4 + x2 − 2 −4 x + 1 c = 2 = x2 + x −1 + 2 d ( x) x − x+3 x − x+3 ⇒ x4 + x2 – 2 = (x2 – x + 3) (x2 + x – 1) + (–4x + 1) giving us q (x) = x2 + x – 1 and r (x) = –4x + 1. Exercise 1.4 1 For each of the following pairs of polynomials, find the quotient q (x) and remainder r (x) that satisfy the requirements of the Polynomial Division Theorem: a f (x) = x2 – x + 7; d (x) = x + 1 b f (x) = x3 + 2x2 – 5x + 3; d (x) = x2 + x – 1 c f (x) = x2 + 8x – 12; d (x) = 2 2 In each of the following, express the function f (x) in the form f (x) = (x – c) q (x) + r (x) for the given number c. 1 a f (x) = x3 – 5x2 – x + 8; c = –2 b f (x) = x3 + 2x2 – 2x – 14; c = 2 3 Perform the following divisions, assuming that n is a positive integer: x3n + 5x2n + 12 xn + 18 x3n − x2n + 3xn − 10 a b xn + 3 xn − 2 21 Mathematics Grade 10 1.2.2 Remainder Theorem The equality f (x) = d (x) q (x) + r (x) expresses the fact that Dividend = (divisor) (quotient) + remainder. ACTIVITY 1.7 1 Let f (x) = x4 – x3 – x2 – x – 2. a Find f (_2) and f (2). b What is the remainder if f (x) is divided by x + 2? c Is the remainder equal to f (–2)? d What is the remainder if f (x) is divided by x – 2? e Is the remainder equal to f (2)? 2 In each of the following, find the remainder when the given polynomial f (x) is divided by the polynomial x – c for the given number c. Also, find f (c). a f (x) = 2x2 + 3x + 1; c = –1 b f (x) = x6 + 1; c = –1, 1 c f (x) = 3x3 – x4 + 2; c = 2 d f (x) = x3 – x + 1; c = –1, 1 Theorem 1.2 Remainder theorem Let f(x) be a polynomial of degree greater than or equal to 1 and let c be any real number. If f(x) is divided by the linear polynomial (x – c), then the remainder is f(c). Proof:- When f (x) is divided by x – c, the remainder is always a constant. Why? By the polynomial division theorem, f (x) = (x – c) q (x) + k where k is constant. This equation holds for every real number x. Hence, it holds when x = c. In particular, if you let x = c, observe a very interesting and useful relationship: f (c) = (c – c) q (c) + k = 0. q (c) + k =0+k=k It follows that the value of the polynomial f (x) at x = c is the same as the remainder k obtained when you divide f (x) by x – c. 22  Unit 1 Polynomial Functions Example 2 Find the remainder by dividing f (x) by d (x) in each of the following pairs of polynomials, using the polynomial division theorem and the remainder theorem: a f (x) = x3 – x2 + 8x – 1; d (x) = x + 2 b f (x) = x4 + x2 + 2x + 5; d (x) = x – 1 Solution: a Polynomial division theorem Remainder theorem x3 − x 2 + 8 x − 1 3 2 f ( − 2 ) = ( − 2 ) − ( − 2 ) + 8 ( − 2 ) − 1, x+2 29 = x2 − 3x + 14 − = –8 – 4 – 16 – 1 = –29 x+2 Therefore, the remainder is –29. b Polynomial division theorem Remainder theorem x4 + x2 + 2x + 5 f (1) = (1)4 + (1)2 + 2(1) + 5 x −1 9 = x3 + x2 + 2x + 4 + =1+1+2+5=9 x − 1 Therefore, the remainder is 9. Example 3 When x3 – 2x2 + 3bx + 10 is divided by x – 3 the remainder is 37. Find the value of b. Solution: Let f (x) = x3 – 2x2 + 3bx + 10. f (3) = 37. (By the remainder theorem) ⇒ (3)3 – 2 (3)2 + 3b(3) + 10 = 37 27 – 18 + 9b + 10 = 37 ⇒ 9b + 19 = 37 ⇒ b = 2. Exercise 1.5 1 In each of the following, express the function in the form f (x) = (x – c) q (x) + r (x) for the given number c, and show that f (c) = k is the remainder. a f (x) = x3 – x2 + 7x + 11; c = 2 b f (x) = 1 – x5 + 2x3 + x; c = –1 2 c f (x) = x4 + 2x3 + 5x2 + 1; c = − 3 23 Mathematics Grade 10 2 In each of the following, use the Remainder Theorem to find the remainder k when the polynomial f (x) is divided by x – c for the given number c. 1 a f (x) = x17 – 1; c = 1 b f (x) = 2x2 + 3x + 1; c = − 2 c f (x) = x23 + 1; c = –1 3 When f (x) = 3x7 – ax6 + 5x3 – x + 11 is divided by x + 1, the remainder is 15. What is the value of a? 4 When the polynomial f (x) = ax3 + bx2 – 2x + 8 is divided by x – 1 and x + 1 the remainders are 3 and 5 respectively. Find the values of a and b. 1.2.3 Factor Theorem Recall that, factorizing a polynomial means writing it as a product of two or more polynomials. You will discuss below an interesting theorem, known as the factor theorem, which is helpful in checking whether a linear polynomial is a factor of a given polynomial or not. ACTIVITY 1.8 1 Let f (x) = x3 – 5x2 + 2x + 8. a Find f (2). b What is the remainder when f (x) is divided by x – 2? c Is x – 2 a factor of f (x)? d Find f (–1) and f (1). e Express f (x) as f (x) = (x – c) q (x) where q (x) is the quotient. 2 Let f (x) = x3 – 3x2 – x + 3. a What are the values of f (–1), f (1) and f (3)? b What does this tell us about the remainder when f (x) is divided by x + 1, x – 1 and x – 3? c How can this help us in factorizing f (x)? Theorem 1.3 Factor theorem Let f(x) be a polynomial of degree greater than or equal to one, and let c be any real number, then i x – c is a factor of f(x), if f(c) = 0, and ii f(c) = 0, if x – c is a factor of f(x). Try to develop a proof of this theorem using the remainder theorem. 24  Unit 1 Polynomial Functions Group Work 1.2 4 2 1 Let f ( x) = 4x − 5x + 1. a Find f (–1) and show that x + 1 is a factor of f (x). b Show that 2x – 1 is a factor of f (x). c Try to completely factorize f (x) into linear factors. 2 Give the proof of the factor theorem. Hint: You have to prove that i if f (c) = 0, then x – c is a factor of f (x) ii if x – c is a factor of f (x), then f (c) = 0 Use the polynomial division theorem with factor (x – c) to express f (x) as f (x) = d (x) q (x) + r (x), where d (x) = x – c. Use the remainder theorem r (x) = k = f (c), giving you f (x) = (x – c) q (x) + f (c) where q (x) is a polynomial of degree less than the degree of f (x). If f (c) = 0, then what will f (x) be? Complete the proof. Example 4 Let f (x) = x3 + 2x2 – 5x – 6. Use the factor theorem to determine whether: a x + 1 is a factor of f (x) b x + 2 is a factor of f (x). Solution: a Since x + 1 = x – (–1), it has the form x – c with c = –1. f (–1) = (–1)3 + 2(–1)2 – 5(–1) – 6 = –1 + 2 + 5 – 6 = 0. So, by the factor theorem, x + 1 is a factor of f (x). b f (–2) = (–2)3 + 2 (–2)2 – 5 (–2) – 6 = –8 + 8 + 10 – 6 = 4 ≠ 0. By the factor theorem, x + 2 is not a factor of f (x). Example 5 Show that x + 3, x – 2 and x + 1 are factors and x + 2 is not a factor of f (x) = x4 + x3 – 7x2 – x + 6. Solution: f (–3) = (–3)4 + (–3)3 – 7(–3)2 – (–3) + 6 = 81 – 27 – 63 + 3 + 6 = 0. Hence x + 3 is a factor of f (x). f (2) = 24 + (2)3 – 7 (2)2 – 2 + 6 = 16 + 8 – 28 – 2 + 6 = 0. 25 Mathematics Grade 10 Hence x – 2 is a factor of f (x). f (–1) = (–1)4 + (–1)3 – 7 (–1)2 – (–1) + 6 = 1 – 1 – 7 + 1 + 6 = 0 Hence x + 1 is a factor of f (x). f (–2) = (–2)4 + (–2)3 – 7 (–2)2 – (–2) + 6 = 16 – 8 – 28 + 2 + 6 = –12 ≠ 0 Hence x + 2 is not a factor of f (x). Exercise 1.6 1 In each of the following, use the factor theorem to determine whether or not g (x) is a factor of f (x). a g (x) = x +1; f (x) = x15 +1 b g (x) = x −1; f (x) = x7 + x −1 3 c g ( x) = x − ; f ( x) = 6 x 2 + x − 1 2 d g ( x) = x + 2; f ( x) = x 3 − 3x 2 − 4 x − 12 2 In each of the following, find a number k satisfying the given condition: a x − 2 is a factor of 3 x 4 − 8 x 2 − kx + 6 b x + 3 is a factor of x5 − kx 4 − 6 x3 − x 2 + 4 x + 29 c 3x − 2 is a factor of 6 x3 − 4 x 2 + kx − k 3 Find numbers a and k so that x – 2 is a factor of f (x) = x4 – 2ax3 + ax2 – x + k and f (–1) = 3. 4 Find a polynomial function of degree 3 such that f (2) = 24 and x – 1, x and x + 2 are factors of the polynomial. 5 Let a be a real number and n a positive integer. Show that x – a is a factor of xn – an. 6 Show that x – 1 and x + 1 are factors and x is not a factor of 2x3 – x2 – 2x + 1. 7 In each of the following, find the constant c such that the denominator will divide the numerator exactly: x3 + 3 x 2 − 3 x + c x3 − 2 x 2 + x + c a b. x −3 x+2 2 8 The area of a rectangle in square feet is x + 13x + 36. How much longer is the length than the width of the rectangle? 26  Unit 1 Polynomial Functions 1.3 ZEROS OF A POLYNOMIAL FUNCTION In this section, you will discuss an interesting concept known as zeros of a polynomial. Consider the polynomial function f (x) = x – 1. What is f (1)? Note that f (1) = 1 – 1 = 0. As f (1) = 0, we say that 1 is the zero of the polynomial function f (x). To find the zero of a linear (first degree polynomial) function of the form f (x) = ax + b, a ≠ 0, we find the number x for which ax + b = 0. Note that every linear function has exactly one zero. ax + b = 0 ⇒ ax = –b........ Subtracting b from both sides b ⇒ x= −........ Dividing both sides by a, since a ≠ 0. a b Therefore, x = − is the only zero of the linear function f, whenever a ≠ 0. a 2x −1 x + 2 Example 1 Find the zeros of the polynomial f ( x ) = − −2. 3 3 2x −1 x + 2 Solution: f ( x) = 0 ⇒ − =2 3 3 2x − 1−(x + 2) = 6 ⇒ 2x − 1− x − 2 = 6 ⇒ x = 9. So, 9 is the zero of f (x). Similarly, to find the zeros of a quadratic function (second degree polynomial) of the form f (x) = ax2 + bx + c, a ≠ 0, we find the number x for which ax2 + bx + c = 0, a ≠ 0. ACTIVITY 1.9 1 Find the zeros of each of the following functions: 3 a h (x) = 1 – (x + 2) b k (x) = 2 – (x2 – 4) + x2 – 4x 5 c f (x) = 4x2 – 25 d f (x) = x2 + x – 12 e f (x) = x3 – 2x2 + x f g (x) = x3 + x2 – x –1 2 How many zeros can a quadratic function have? 3 State techniques for finding zeros of a quadratic function. 4 How many zeros can a polynomial function of degree 3 have? What about degree 4? Example 2 Find the zeros of each of the following quadratic functions: a f (x) = x2 – 16 b g (x) = x2 – x – 6 c h (x) = 4x2 – 7x + 3 27 Mathematics Grade 10 Solution: a f (x) = 0 ⇒ x2 – 16 = 0 ⇒ x2 – 42 = 0 ⇒ (x – 4) (x + 4) = 0 ⇒ x – 4 = 0 or x + 4 = 0 ⇒ x = 4 or x = – 4 Therefore, – 4 and 4 are the zeros of f. b g (x) = 0 ⇒ x2 – x – 6 = 0 Find two numbers whose sum is – 1 and whose product is – 6. These are – 3 and 2. x2 – 3x + 2x – 6 = 0 ⇒ x (x – 3) + 2 (x – 3) = 0 ⇒ (x + 2) (x – 3) = 0 ⇒ x + 2 = 0 or x – 3 = 0 ⇒ x = –2 or x = 3 Therefore, –2 and 3 are the zeros of g. c h (x) = 0 ⇒ 4x2 – 7x + 3 = 0 Find two numbers whose sum is –7 and whose product is 12. These are –4 and –3. Hence, 4x2 – 7x + 3 = 0 ⇒ 4x2 – 4x – 3x + 3 = 0 ⇒ 4x (x – 1) – 3 (x – 1) = 0 3 ⇒ (4x – 3) (x – 1) = 0 ⇒ 4x – 3 = 0 or x – 1 = 0 ⇒ x = or x = 1. 4 3 Therefore, and 1 are the zeros of h. 4 Definition 1.2 For a polynomial function f and a real number c, if f(c) = 0, then c is a zero of f. Note that if x – c is a factor of f (x), then c is a zero of f (x). Example 3 a Use the factor theorem to show that x + 1 is a factor of f (x) = x25 + 1. b What are the zeros of f (x) = 3 (x – 5) (x + 2) (x – 1)? c What are the real zeros of x4 – 1 = 0? d Determine the zeros of f (x) = 2x4 – 3x2 + 1. Solution: a Since x + 1 = x – (–1), we have c = –1 and f (c) = f (–1) = (–1)25 + 1 = –1 + 1 = 0 Hence, –1 is a zero of f (x) = x25 + 1, by the factor theorem. So, x – (–1) = x + 1 is a factor of x25 + 1. b Since (x – 5), (x + 2) and (x – 1) are all factors of f (x), 5, –2 and 1 are the zeros of f (x). 28  Unit 1 Polynomial Functions c Factorising the left side, we have x4 – 1 = 0 ⇒ (x2 – 1) (x2 + 1) = 0 ⇒ (x – 1) (x + 1) (x2 + 1) = 0 So, the real zeros of f (x) = x4 – 1 are –1 and 1. d f (x) = 0 ⇒ 2x4 – 3x2 + 1 = 0 ⇒ 2 (x2)2 – 3x2 + 1 = 0 Let y = x2. Then 2(y)2 – 3y + 1 = 0 ⇒ 2y2 – 3y + 1 = 0 ⇒ (2y – 1) (y – 1) = 0 ⇒ 2y – 1 = 0 or y – 1 = 0 1 Hence y = or y = 1 2 1 Since y = x2, we have x2 = or x2 = 1. 2 1 1 2 Therefore x = ± or x = ± 1. (Note that =.) 2 2 2 2 2 Hence, − , , –1 and 1 are zeros of f. 2 2 A polynomial function cannot have more zeros than its degree. 1.3.1 Zeros and Their Multiplicities If f (x) is a polynomial function of degree n, n ≥ 1, then a root of the equation f (x) = 0 is called a zero of f. By the factor theorem, each zero c of a polynomial function f (x) generates a first degree factor (x – c) of f (x). When f (x) is factorized completely, the same factor (x – c) may occur more than once, in which case c is called a repeated or a multiple zero of f (x). If x – c occurs only once, then c is called a simple zero of f (x). Definition 1.3 If (x – c)k is a factor of f (x), but (x – c)k+1 is not, then c is said to be a zero of multiplicity k of f. Example 4 Given that –1 and 2 are zeros of f (x) = x4 + x3 − 3x2 − 5x − 2, determine their multiplicity. Solution: By the factor theorem, (x + 1) and (x − 2) are factors of f (x) Hence, f (x) can be divided by (x + 1) (x − 2) = x2 − x − 2, giving you f (x) = (x2 − x − 2) (x2 + 2x + 1) = (x + 1) (x − 2) (x + 1)2 = (x + 1)3 (x − 2) Therefore, –1 is a zero of multiplicity 3 and 2 is a zero of multiplicity 1. 29 Mathematics Grade 10 Exercise 1.7 1 Find the zeros of each of the following functions: 3 1 a f (x) = 1 – x b f (x) = (1 – 2x) – (x + 3) 5 4 2 c g (x) = (2 – 3x) (x – 2) (x + 1) d h (x) = x4 + 7x2 + 12 3 e g (x) = x3 + x2 – 2 f f (t) = t3 – 7t + 6 g f (y) = y5 – 2y3 + y h f (x) = 6x4 – 7x2 – 3 2 For each of the following, list the zeros of the given polynomial and state the multiplicity of each zero.  2 2 a f (x) = x12  x −   3 b ( g (x) = 3 x − 2 ) ( x + 1) 5 c 5 h (x) = 3 x 6 ( π − x ) ( x − ( π + 1) ) 3 d ( f ( x) = 2 x − 3 ) ( x + 5) (1 − 3x ) 9 3 2 e f (x) = x − 3x + 3x −1 3 Find a polynomial function f of degree 3 such that f (10) = 17 and the zeros of f are 0, 5 and 8. 4 In each of the following, the indicated number is a zero of the polynomial function f (x). Determine the multiplicity of this zero. a 1; f (x) = x3 + x2 – 5x + 3 b –1; f (x) = x4 + 3x3 + 3x2 + x 1 c ; f ( x) = 4 x3 − 4 x2 + x. 2 4 5 Show that if 3x + 4 is a factor of some polynomial function f, then − is a zero of f. 3 6 In each of the following, find a polynomial function that has the given zeros satisfying the given condition. a 0, 3, 4 and f (1) = 5 b −1, 1 + 2, 1 − 2 and f (0) = 3. 1 7 A polynomial function f of degree 3 has zeros −2, and 3, and its leading 2 coefficient is negative. Write an expression for f. How many different polynomial functions are possible for f ? 30  Unit 1 Polynomial Functions 8 If p (x) is a polynomial of degree 3 with p (0) = p (1) = p (–1) = 0 and p (2) = 6, then a show that p (–x) = –p (x). b find the interval in which p (x) is less than zero. 9 Find the values of p and q if x – 1 is a common factor of f ( x ) = x 4 − px 3 + 7 qx + 1, and g ( x ) = x 6 − 4 x 3 + px 2 + qx − 3. 10 The height above ground level in metres of a missile launched vertically, is given by h(t ) = −16t 3 +100t. At what time is the missile 72 m above ground level? [t is time in seconds]. 1.3.2 Location Theorem A polynomial function with rational coefficients may have no rational zeros. For example, the zeros of the polynomial function: f (x) = x2 – 4x – 2 are all irrational. Can you work out what the zeros are? The polynomial function p (x) = x3 – x2 – 2x + 2 has rational and irrational zeros, − 2,1 and 2. Can you check this? ACTIVITY 1.10 1 In each of the following, determine whether the zeros of the corresponding function are rational, irrational, or neither: a f (x) = x2 + 2x + 2 b f (x) = x3+ x2 – 2x – 2 c f (x) = (x + 1)(2x2 + x – 3) d f (x) = x4 – 5x2 + 6 2 For each of the following polynomials make a table of values, for – 4 ≤ x ≤ 4: a f (x) = 3x3 + x2 + x – 2 b f (x) = x4 – 6x3 + x2 + 12x – 6 Most of the standard methods for finding the irrational zeros of a polynomial function involve a technique of successive approximation. One of the methods is based on the idea of change of sign of a function. Consequently, the following theorem is given. Theorem 1.4 Location theorem Let a and b be real numbers such that a < b. If f is a polynomial function such that f(a) and f(b) have opposite signs, then there is at least one zero of f between a and b. This theorem helps us to locate the real zeros of a polynomial function. It is sometimes possible to estimate the zeros of a polynomial function from a table of values. 31 Mathematics Grade 10 Example 5 Let f (x) = x4 – 6x3 + x2 + 12x – 6. Construct a table of values and use the location theorem to locate the zeros of f between successive integers. Solution: Construct a table and look for changes in sign as follows: x –3 –2 –1 0 1 2 3 4 5 6 f (x) 210 38 –10 –6 2 –10 –42 –70 –44 102 Since f (–2) = 38 > 0 and f (–1) = –10 < 0, we see that the value of f (x) changes from positive to negative between – 2 and –1. Hence, by the location theorem, there is a zero of f (x) between x = –2 and x = –1. Since f (0) = – 6 < 0 and f (1) = 2 > 0, there is also one zero between x = 0 and x = 1. Similarly, there are zeros between x = 1 and x = 2 and between x = 5 and x = 6. Example 6 Using the location theorem, show that the polynomial f (x) = x5 – 2x2 – 1 has a zero between x = 1 and x = 2. Solution: f (1) = (1)5 – 2(1)2 – 1 = 1 – 2 – 1 = – 2 < 0. f (2) = (2)5 – 2 (2)2 – 1 = 32 – 8 – 1 = 23 > 0. Here, f (1) is negative and f (2) is positive. Therefore, there is a zero between x = 1 and x = 2. Exercise 1.8 1 In each of the following, use the table of values for the polynomial function f (x) to locate zeros of y = f (x): a x –5 –3 –1 0 2 5 f (x) 7 4 2 –1 3 –6 b x –6 –5 –4 –3 –2 –1 0 1 2 f (x) –21 –10 8 –1 –5 6 4 –3 18 2 Use the location theorem to verify that f (x) has a zero between a and b: a f (x) = 3x3 + 7x2 + 3x + 7 ; a = –3 , b = –2 3 b f (x) = 4x4 + 7x3 – 11x2 + 7x – 15; a = 1, b = 2 c f (x) = –x4 + x3 + 1 ; a = –1 , b = 1 d f (x) = x5 – 2x3 – 1 ; a = 1, b = 2 32  Unit 1 Polynomial Functions 3 In each of the following, use the Location Theorem to locate each real zero of f (x) between successive integers: a f (x) = x3 – 9x2 + 23x – 14; for 0 ≤ x ≤ 6 b f (x) = x3 – 12x2 + x + 2; for 0 ≤ x ≤ 8 c f (x) = x4 – x2 + x – 1; for –3 ≤ x ≤ 3 d f (x) = x4 + x3 – x2 – 11x + 3; for –3 ≤ x ≤ 3 4 In each of the following, find all real zeros of the polynomial function, for −4 ≤ x ≤ 4 : 15 2 a f (x) = x4 – 5x3 + x − 2x − 2 b f (x) = x5 – 2x4 – 3x3 + 6x2 + 2x – 4 2 c f (x) = x + x – 4x2 – 2x + 4 4 3 d f (x) = 2x4 + x3 – 10x2 – 5x 5 In Question 10 of Exercise 1.7, at what time is the missile 50 m above the ground level? 6 Is it possible for a polynomial function of degree 3 with integer coefficient to have no real zeros? Explain your answer. [ 1.3.3 Rational Root Test The rational root test relates the possible rational zeros of a polynomial with integer coefficients to the leading coefficient and to the constant term of the polynomial. Theorem 1.5 Rational root test p If the rational number , in its lowest terms, is a zero of the polynomial q f (x) = anxn + an–1 xn –1 +.... + a1x + ao with integer coefficients, then p must be a factor of ao and q must be a factor of an. ACTIVITY 1.11 1 What should you do first to use the rational root test? 2 What must the leading coefficient be for the possible rational zeros to be factors of the constant term? 3 Suppose that all of the coefficients are rational numbers. What could be done to change the polynomial into one with integer coefficients? Does the resulting polynomial have the same zeros as the original? 4 There is at least one rational zero of a polynomial whose constant term is zero. What is this number? 33 Mathematics Grade 10 Example 7 In each of the following, find all the rational zeros of the polynomial: a f (x) = x3 – x + 1 b g (x) = 2x3 + 9x2 + 7x – 6 1 1 c g ( x) = x 4 − 2 x3 − x 2 + 2 x 2 2 Solution: a The leading coefficient is 1 and the constant term is 1. Hence, as these are factors of the constant term, the possible rational zeros are ± 1. Using the remainder theorem, test these possible zeros. f (1) = (1)3 – 1 + 1 = 1 – 1 + 1 = 1 f (–1) = (–1)3 – (–1) + 1 = –1 + 1 + 1 = 1 So, we can conclude that the given polynomial has no rational zeros. b an = a3 = 2 and ao = –6 Possible values of p are factors of –6. These are ±1, ±2, ±3 and ±6. Possible values of q are factors of 2. These are ± 1, ± 2. p 1 3 The possible rational zeros are ± 1, ± 2, ± 3, ± 6, ± , ±. q 2 2 Of these 12 possible rational zeros, at most 3 can be the zeros of g (Why?). 1 Check that f (–3) = 0, f (– 2) = 0 and f   = 0. 2 Using the factor theorem, we can factorize g (x) as: 2x3 + 9x2 + 7x – 6 = (x + 3) (x + 2) (2x – 1). So, g (x) = 0 at 1 x = –3, x = –2 and at x =. 2 1 Therefore –3, –2 and are the only (rational) zeros of g. 2 c Let h (x) = 2g(x). Thus h (x) will have the same zeros, but has integer coefficients. h (x) = x4 – 4x3 – x2 + 4x x is a factor, so h (x) = x(x3 – 4x2 – x + 4) = xk (x) k (x) has a constant term of 4 and leading coefficient of 1. The possible rational zeros are ±1, ±2, ±4. 34  Unit 1 Polynomial Functions Using the remainder theorem, k (1) = 0, k (–1) = 0 and k (4) = 0 So, by the factor theorem k (x) = (x – 1) (x + 1) (x – 4). Hence, h (x) = x k (x) = x(x – 1) (x + 1) (x – 4) and 1 1 g (x) = h( x) = x(x – 1)(x + 1)(x – 4). 2 2 Therefore, the zeros of g (x) are 0, ± 1 and 4. Exercise 1.9 1 In each of the following, find the zeros and indicate the multiplicity of each zero. What is the degree of the polynomial? a f (x) = (x + 6) (x – 3)2 b f (x) = 3 (x + 2)3 (x – 1)2 (x + 3) 1 c f (x) = (x – 2)4 (x + 3)3 (1 – x) d f (x) = x4 – 5x3 + 9x2 – 7x + 2 2 e f (x) = x4 – 4x3 + 7x2 – 12x + 12 2 For each of the following polynomials, find all possible rational zeros: a p (x) = x3 – 2x2 – 5x + 6 b p (x) = x3 – 3x2 + 6x + 8 c p (x) = 3x3 – 11x2 + 8x + 4 d p (x) = 2x3 + x2 – 4x – 3 e p (x) = 12x3 – 16x2 – 5x + 3 3 In each of the following, find all the rational zeros of the polynomial, and express the polynomial in factorized form: a f (x) = x3 − 5x2 − x + 5 b g (x) = 3x3 + 3x2 − x −1 c p (t ) = t 4 − t 3 − t 2 − t − 2 4 In each of the following, find all rational zeros of the function: 3 11 2 1 1 4 25 2 a p (y) = y + y − y− b p (x) = x − x +9 6 2 3 4 4 21 2 3 4 7 3 7 2 5 c h (x) = x − x + x d p (x) = x + x − x − x 10 5 6 3 2 5 For each of the following, find all rational roots of the polynomial equation: a 2x3 – 5x2 + 1 = 0 b 4x4 + 4x3 – 9x2 – x + 2 = 0 c 2x5 – 3x4 – 2x + 3 = 0 35 Mathematics Grade 10 1.4 GRAPHS OF POLYNOMIAL FUNCTIONS In your previous grades, you have discussed how to draw graphs of functions of degree zero, one and two. In the present section, you will learn about graphs of polynomial functions of degree greater than two. To understand properties of polynomial functions, try the following Activity. ACTIVITY 1.12 1 Sketch the graph of each of the following polynomial functions: a f (x) = 3 b f (x) = – 2.5 c g (x) = x – 2 d g (x) = –3x + 1 2 Let f (x) = x2 –4x +5 a Copy and complete the table of values given below. x –2 –1 0 1 2 3 4 2 f (x) = x – 4x + 5 b Plot the points with coordinates (x, y), where y = f (x) on the xy-coordinate plane. c Join the points in b above by a smooth curve to get the graph of f. What do you call the graph of f ? Give the domain and range of f. 3 Construct a table of values for each of the following polynomial functions and sketch the graph: a f (x) = x2 − 3 b g (x) = −x2 − 2x +1 c h (x) = x3 d p (x) = 1− x4 We shall discuss sketching the graphs of higher degree polynomial functions through the following examples. 3 Example 1 Let us consider the function p ( x) = x − 3x − 4. 3 This function can be written as y = x − 3x − 4 Copy and complete the table of values below. x –3 –2 –1 0 1 2 3 y –6 –2 –6 14 Other points between integers may help you to determine the shape of the graph better. 36  Unit 1 Polynomial Functions 1 For instance, for x = 2 1 43 y = p  = − 2 8  1 43  Therefore, the point  , −  is on the graph of p. Similarly, for 2 8  5  5  33 x = , y= p  =. 2  2 8  5 33  So,  ,  is also on the graph of p. 2 8  Plot the points with coordinates (x, y) from the table as shown in Figure 1.3a. Now join these points by a smooth curve to get the graph of p (x), as shown in Figure 1.3b. y y 10 10 5 5 x x -5 5 -5 5 -5 -5 -10 -10 a b Figure 1.3 Graph of p (x) = x3 − 3x − 4 Example 2 Sketch the graph of f ( x) = − x 4 + 2 x2 + 1 Solution: To sketch the graph of f, we find points on the graph using a table of values. x –2 –1 0 1 2 y = −x4 + 2x2 + 1 –7 2 1 2 –7 Plot the points with coordinates (x, y) from this table and join them by a smooth curve for increasing values of x, as shown in Figure 1.4. 37 Mathematics Grade 10 From the graph, find the domain and the y range of f. Observe that the graph of f opens 3 downward. 2 As observed from the above two examples, 1 x the graph of a polynomial function has no -2 -1 1 2 3 -1 jumps, gaps and holes. It has no sharp -2 corners. The gra