MATHEMATICS N3 NEW SYLLABUS

Transcription

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2 MATHEMATICS N3 NEW SYLLABUS

3 MATHEMATICS N3 New Syllabus MJJ van Rensburg TROUPANT Publishers

4 Copyright 1994 by the author All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means without prior written permission by the publisher. ISBN: ; eisbn: First edition 1998 Eleventh impression 2013 Published by: Troupant Publishers P.O. Box 4532 Northcliff 2115 Previously published by Southern Book Publishers First edition 1994 Cover design by Alix Gracie Set in 10.5 on 12pt Times New Roman Typesetting by Unifoto (Pty) Ltd, Cape Town Printed and bound by Ultra Litho (Pty) Limited

5 Preface The accent in this book has been put on understanding and not necessarily on formal deductions or rote methods. Examples have been written out fully for students to follow reasoning easily. Module 0 contains revision of basic Mathematics and can be use as orientation. At the end ofthe book there is a summary to help students to structure their own framework of reference. To know about something does not necessarily mean that one can do it, therefore normal exercises are included as well as criterium tests at the back ofthe book. These tests can be used for additional exercise or as evaluation tests at the end of each module. They can also be used for revision if old examination papers are not available. THE AUTHOR

6 Contents Module 0: Orientation 0.1 The importance of terminology 0.2 Symbols 0.3 The operations, X, -;-, + and Variables 0.5 Coefficients 0.6 Examples of terms 0.7 Similar terms 0.8 Addition and subtraction ofalgebraic terms Exercise Laws for exponents 6 Exercise Fractions Multiplication and division by fractions Factors Simplification of fractions Addition and subtraction of fractions Exercise The pocket calculator Powers The memory Calculations with parentheses (brackets) Exercise 0.4

7 Module 1: Factors and fractions 1.1 The common factor Ge.neral Common factors Grouping Exercise The quadratic trinomial General The solvability of trinomials The traditional method Exercise Squares The difference between two squares Completing the square Exercise Cube functions Function notation Function values The remainder theorem Exercise Factors with the aid of the remainder theorem 40 Exercise Algebraic fractions Simplification of fractions Multiplication of fractions Division of fractions Exercise Addition and subtraction of fractions 49 Exercise 1.7

8 Module 2: Exponents, surds and logarithms 2.1 Exponents General Exponent laws Exercise Exponential equations Exercise Surds General Multiplication with equal order Changing the order Writing mixed surds as one surd Addition and subtraction of surds Multiplication and division of surds Rationalising denominators Factors Exercise Equations containing surds 78 Exercise Logarithms The meaning ofa logarithm Calculations with respect to bases, numbers and logarithms, without using a calculator Ex,ercise Logarithms and anti-logarithms using a calculator 83 Exercise Logarithmic laws Simplifications without using a calculator Exercise Changing the base 89 Exercise 2.8

9 Module 3: Equations, word problems and manipulation of technical formulae Equations Solving a linear equation Exercise Simultaneous linear equations 96 Exercise Solving simultaneous equations where one is linear and the other quadratic 101 Exercise Word problems Compiling and solving simultaneous equations Exercise Compiling and solving quadratic equations 106 Exercise Manipulation of technical fo(mulae Introduction The most elementary 'formula' Different operations in the same equation Exercise Additional examples 116 Exercise 3.7 Module 4: Geometry of co-ordinates 4.1 Different forms ofa straight line The gradient-intercept form The intercept form

10 4.1.3 The general form Ifthe angle between the positive x-axis and the line is given Ifthe equation is given Iftwo points are given Parallel lines Perpendicular lines 128 Exercise Equations of straight lines Iftwo points are given If one point and a gradient are given The distance between two points The co-ordinates of the midpoint of a line segment 138 Exercise The circle The equation ofa circle with its centre at the origin The intersections ofa straight line and a circle The equation of a tangent to a circle in a given direction The equation of a tangent at a given point on the circumference ofa circle 148 Exercise 4.3 Module 5: Sketch graphs General The straight line The gradient-intercept form The general form The intercept form A circle with its centre at the origin The ellipse The parabola The hyperbola The cube function The form of y = ax n Exercise 5

11 Module 6: Differential calculus 6.1 Average gradient and speed The average gradient Average speed Exercise Limits General The gradient of a tangent to a curve at a point Exercise Differentiation General Rules of differentiation Exercise The gradient of a curve at a point on the curve 191 Exercise Turning points 193 Exercise 6.5 Module 7: Trigonometry Exact values General Exact values in the first quadrant Exact values in other quadrants Another rhyme to help you remember signs Solving easy trigonometric equations with the aid of exact values Exercise Trigonometric equations Using the calculator to determine function and arc function values The solution of 'easy' trigonometric equations More advanced trigonometric equations 218

12 Exercise Identities 223 Exercise Solving triangles Right-angled triangles The sine rule The cosine rule Exercise The area of a triangle Exercise Sketch graphs Periodic functions Exercise Superimposing graphs 255 Exercise Rotating vectors Sketching trigonometric graphs from a rotating vector Superimposing graphs Graphs of the form y = sin k () Frequency 262 Exercise Summary 266 Criterium tests 267

13 Module 0 Orientation Objectives and overview This module gives an overview ofthe pre-knowledge required by the student. On completion of this section, you should be able to: 1. Use the four fundamental processes +, -, x and -;- with and without a pocket calculator. 2. State the difference between terms and factors. 3. Add and subtract simple algebraic terms. 4. Reproduce the index laws and apply them in simple calculations. 5. Multiply, divide, add and subtract simple fractions. 6. Use a calculator in simple calculations. 0.1 The importance of terminology Notation is very important. If you don't know the Mathematical language, you will not be able to: Read and understand Mathematics. Solve mathematical problems. Write down the correct solutions. Find mistakes in solutions. 0.2 Symbols is equal to + plus minus divide therefore because

14 -::t= is not equal to > is greater than (e.g. 6 > 4) < is smaller than (e.g. 3 < 5) ~ is greater or equal to ~ is smaller or equal to p is not greater than \: is not smaller than ~ strives to 00 infinity => implicates ex proportional to e 2, circumference of a circle diameter of a circle is parallel to.::!:- is perpendicular to AB the length of the line between A and B ~ implicates and being implicated 0.3 The operations, x, + and - + x + + x + + x x + When these operations occur in the same sentence, we must first x and -7- and then + and -, but always work from left to right. Examples: x 5 -: x x 6 Solutions: x = :- 2 = = -10 2

15 With a calculator: rn[±]q]e]@]0rnelrng Beware of: rn[±]q]ge]@]g0rngelrng This upsets the sequence ;- 4 x = 0,5 x = 1, = -2,5 Press rn El [i] 0 Q] [±] rn G Beware of: 2-;-4x3-6+2 "* 2 -; *~-6+2 * -3~ You must work from left to right. First -;- and then x. x and -;- both have the same status ; x 6 4 =2-"3+ 2x6 4 = 2 - " = 12j Press rn E] [i] El Q] [±] rn 0.4 Variables Variables are very important in Mathematics. You started to use them very early in school, e.g. 3 + D = 5. The D represents an unknown which we also call a variable. The same equation can also be written as 3 + x = 5, where x represents the variable. When there are two unknowns in an equation we need the term variable. For example: x - y = 2. This equation is true if x and y represent several different values: 2-0 = 2 3-1=2 4-2 = 2. 3 G

16 It is clear that the 'unknowns' can take on many values, that is why we prefer to call them variables. Variables are indicated by letters, e.g. a, b, c, x, y, z etc. Greek letters can also be used, e.g. a, {3, r, 9, qj etc. 0.5 Coefficients If a number is multiplied by a variable, the number is called the coefficient, e.g. 3a = 3 x a, 3 is the coefficient and a is the variable or symbol. The meaning of 3a is 3 of a, or a + a + a. Another example is: 4x = 4 x x which is equal to x + x + x + x. If the coefficient is I we don't write it down, e.g. a instead of la. In 3a 2 it follows that: 3 is called the coefficient a is called the base 2 is called the index (exponent) a 2 is called the power 0.6 Examples of terms Terms are separated by + and/or - in an equation or expression. For example: a + b 2 terms (binomial) a+b-c 3 terms (trinomial) 2a + 4b = 8c 3 terms in an equation 2a - 4b + 6a - 8ab more than 3 terms (polynomial). 0.7 Similar terms Terms where only the coefficients differ are called similar terms, e.g. a, 2a, 6a, ~a or xy, 3xy, 12xy, -~xy or 2x 2, 4x 2, ~X2 etc. Non-similar terms will be: 2a, 3b or 2x, 2x 2, 3x 3 etc. 0.8 Addition and subtraction of algebraic terms Only similar terms can be added or subtracted. Example: 6a + 2a - 4a = ( )a = 4a [ of a] 4

17 Example: 3a + 3b + 2ab - 6a + ab = 3b + 3a - 6a + 2ab + ab = 3b + (3-6) a + (2 + l)ab = 3b - 3a + 3ab Examples: 1. Add 2a + 3ab - 6b to 2-2b + 4a 2. Subtract 3x + xy - 2 from 2xy + x - 4 Solutions: 1. 2a + 3ab - 6b +4a - 2b + 2 6a + 3ab - 8b x + xy - 2 -x+2xy-4 2x- xy+2 [Rearrange to ensure that similar terms are positioned underneath each other.] [A short method is to change the signs of the terms that must be subtracted and then add.] Exercise 0.1 Simplify: 1. 3a + 2ab - 6a + 2b - ab 2. 2xy + 3x - 2y - 3xy + x 3. a a ab + b - 4ab - 5b 5. 3p + 2q - 4pq - 2qp Add the following expressions: 6. 2a + 3ab; 2a - 3ab 7. a - 2; 2b xy + 3x - y; 4xy - 2x a - 3; b + 4; c - 12 Subtract the second expression from the first: 10. a - 2; 2b a + b; 3b xy + 2x - y; x - 3y + 2xy x - 6y; 4-2x + 3y 5

18 Simplify: 14. (2x + 3y) + (8x - 2y) 15. (2x 2 + 3x - 2) - (x 2 + 2x + 1) x 8-4 -; x 8-6 -; ; x ; x 2 Answers to Exercise ab + 2b - 3a 3. 4a p + 2q - 6pq 7. a + 2b a + b + c a - 2b x - 9y 15. x 2 + x , x - 2y - xy 4. -2ab - 4b 6. 4a 8. 6xy + x - y a - 2b xy + x + 2y 14. lax + y , Laws for exponents am x an = a m + n ao = 1 (a =#= 0) _ a =--;;; a Examples: Simplify: 1. 2a x 3a x 2 x 5x 3 X 2y -;- 3x 4 ( 4X2y3 ) xy po - (238q)O + (12Sq3)3 6

19 Solutions: a x 3a 2 = (2 x 3)a l + 2 = 6a 3 3x 2 X 5x 3 X 2y -;- 3x 4 = (3 x 5 x 2 -;- 3) r + 3 4y 10xy (4X Y Y (26:;2~J xyy (2 2 'X~ 2 (2i Y _ 16x 4 y 8-81 'yj- 'y pO- (238q)O + (12SqJ) (1) (SJqJ)l x "3q.3 x " q Note that exponent laws are only valid: when bases are the same, e.g. a 2 x b 3 = a 2 x b 3 for x, -;- and powers, e.g (a 2 + b }i 2 =F a 2x "i + b 2x "i for negative bases only if the indices are whole numbers, e.g. [( _ 2)3]2 = (_2)3x2 = (_2)6 = 64 7

20 1 = (_2) 3x 3 = -2 excludes the imaginary solutions. Imaginary numbers are in the N4 syllabus and are used in subjects like Electrotechnics. Practical insert 0.1 Solving x 3 = (x 2 )3 = (-8)3 1 X = [(-2)3]3 = -2 This means that we have to apply a fractional index (~) to a negative base. The answer, - 2, is true because - 2 x - 2 x - 2 = - 8 but it excludes two other roots, x = 1 + 1,732i and x = 1-1,732i. These are called imaginary roots, where i 2 = -1 by definition. Test: (1 + 1,732i)3 = (1 + 1,732i)2 (1 + 1,732i) = [1 + 2(1,732i) -3][1 + 1,732z1 = [-2 + 3,464z1[1 + 1,732z1 = - 2-3,464i + 3,464i - 6 = -8 Likewise: (1-1,732i)3 = -8 Exercise 0.2 Simplify: l. 49a 2 x 2~b -;- 8a 2 b 3. 4( -2pq)(3pq) 5. 2pq + 3p2q X 4pq a x 3ab x 4ab 2 4. (xy) (xy) "2 X 4xy2 + (2y)4 8

21 7. ( 8a 2 bc 2 ) 4ab 2 c (2X)3 X (4xy)2 -;- -L x 9. (54X 6y )~ 2X 3 y 4 Answers to Exercise a 4 4. xy 3. _24p 2 q 2 15x y 9. 3x y 0.10 Fractions When simplifying fractions, notation sometimes causes problems. The following have the same numerical meaning in algebra: ~3 =!4 = = 4(-3) = -4(3) = -3 -;- 4 = 3 -;- (-4) = 0,75. Note that we cannot do this in graphical representations. In a fraction ~, a is called the numerator and b the denominator. ~ = a(!) and ~ = 5(!) etc b b Multiplication and division by fractions When multiplying fractions, both the numerators and the denominators must be multiplied. For example: ace ace b x d x ] = bdf or 132 -x-x x 3 x 2 2 x 4 x

22 When dividing fractions we invert the divisor and then multiply. This is because x and -7- are opposite operations. ~-7-. b d a d =-xb e ad - be 0.12 Factors Factors are identified by finding products. A factor is always a product. The following examples will explain the concept. In ab, a is a factor of ab and b is also a factor of abo In abc, a is a factor of abc, b is a factor of abc and e is a factor of abc. In 2(x + 3), 2 is a factor of 2(x + 3) and x + 3 is also a factor of 2(x + 3) Simplification of fractions Method 2: x2x2x3 2x2x2x3x x 3 2 = 23~3 X = 1 X Method 3: ~x~x~x~ ~x~x~x~x3 1-3 This can result in cancellation of terms, e.g ~ + 2 =1= ~

23 Method 4: ~~) =~~3 1-3 Method 1 or 2 is preferable. Method 3 is effective ifwe remember that only factors can be divided into each other. Cancelling is not generally valid and can lead to mistakes. Method 4 becomes useless in algebraic fractions. Example: 8(a - 1)2b 2 e 12a(a - l)e 2 x 2 x 2(a - 1)(a - 1) x b x b x e 2 x 2 x 3 x a(a - 1) x e a-i a-i e =-x-x-x--x--xbxbx a a-i e 2 a-i =lxlx-x--xlxbxbxl 3 a 2b 2 (a - 1) or or 3a 8(a - 1)2b 2 e 12a(a - l)e 2 3 (a - 1)2b 2 e 2 2.3a(a - l)e (a _ 1)2 J.b 2e) ) 3a 2(a - l)b 2 3a 8(a - 1)2b 2 e 12a(a - l)e ~ x ~ x 2 (a - 1)(a---l) x b x b x ~ ~ x ~ x 3 x a(n---l-) ~ 2(a - l)b 2 3a 11

24 o. 14 Addition and subtraction of fractions Only similar fractions can be added or subtracted = l(~) + 3(~) 1 = (1 + 3>6 = 4(~) 4-6 This is how we can write it in a shorter and more abstract form, e.g = Example: a If fractions are not similar they must be made similar. 12

25 Example: =4 x 3+3 x = = 9(/2) + 8(/2) 1 = (9 + 8).12 = 17(/2) = [x by 3 = 1 and 4 = 1] The shorter method which begins by finding the LCM of the denominators is abstract and should only be used by more experienced people. By resolving the original fractions into prime factors it is easy to eliminate common factors between the numerator and denominator. Example: x2x3 2x2x :;:: 2 x 2 x 3 x 2-2 x 2 x 2 x 3 [x by 2 = 1 and 3 = 1] 10 3 = = 10(/2) - 3(/2) 13

26 = (10-3) (/2) 7-12 Example: 3 a (2 + a) b l + b - --a- 3 a (2 + a) = -b-x-b + b - a 3 a a b a (2 + a) b b = -b-x-b x a+ b x b x a- a x b x b 3a alb b 2 (2 + a) = -ab-2 + -ab-2 - ab2 = [3a + a 2 b - b 2 (2 + a){ a~2] 3a + a 2 b - 2b 2 - ab 2 Exercise 0.3 ab 2 Simplify: (a - b)2be (a - b)c a 2a 2 6a 3 7. b x 3b b "7 + "7 - "7 11 ~ ~_(a+b) ~ ~ _ (a - b) ~ + ~ _ (a - e). ab 2 ae be a 2 be ab 2 e x "7 x " ;- 9 x 5 8 3xy...:.- 14xy 2. 12x 2 y. 7x 2 y 10. ~ + ~ - ~ a a a "

27 Answers to Exercise a 1. " b b 8. 8y e + 3b l - alb + abc. ab 2 e (a - b)be " ~ a a +. 6b The pocket calculator Powers The function ~ Examples: Calculate the following using a calculator (2, 4)5 4.~ 5.~ (3, 6) ~(2, 6)7 Solutions: = 16 Press rn ~ [!J B 2. (2, 4)5 = 79,626 Press rn [] [!J ~ ~ B 3. (3, 6)2.4 = 21,633 Press Q] ~ rn [] [!J B 4. ~ = 2,520 Press ~ Q] IT] B 5. ~126 = 3,350 Press ill ~ [!J rn B 6..(/(2,6)' = 3,810 Press rn [E:] mb [E:] ~ rn B The memory 1. Press the number and Ix -+ ml to put a number into the memory. 2. Press IRMI to use the number in the memory. 3. Put zero into the memory to cancel a number in the memory. Examples: 12 x 4 = x 9 = 108 Press ill rn Ix -+ ml ~ [!J B Press IRMI ~ [2J B 15

28 4. If you want to add a number to the memory press the number and 1M +I. For example: Put 15 in the memory by pressing [I rn Ix --+ mi Add 4 to 15 by pressing [±] IM +I Test by pressing IRMI = If a number, e.g. 5, must be subtracted from the memory, press rn [I] 1M +I ([I] changes the sign of the number). Test by pressing ~!!1 = 10. Examples: Use a calculator to calculate: Solution: Press rn 0 rn D rn G, put it in the memory by pressing Ix --+ mi Press ~ D [±] G D rn B, add it to the memory by pressing IM+I Press 0 rn ITJ G, subtract it from the memory by pressing [I] IM +I Press IRMI to obtain the answer 47, ~ - (12)0 J +..)2,4 + 11,6 Solution: Press ~ G rn G Ix --+ ml Press [I rn 0 D rn G and subtract from the memory by pressing [I] IM +I Press rn D [±] [±] [I [I G I"' ~I and add to the memory by pressing IM +I Answer IRMI = 2,68 16

29 3. [(2,1)16 + li 2 - W]6 Solution: Press rn D IT] ~ ~ G Ix --+ mi Press IT] ~ D rn EJ [!] G 1M +I D [!] [EJ ~ GJ G 0 IM'I Press IRMI G Answer = , Calculations with parentheses (brackets) Parentheses are used when we wish to perform calculations in an order other than that usually followed for the x, -;-, + and - operations. The key IT] forces the operations prior to it to be pending while the calculations inside the brackets are performed. Pressing a number and then IT] gives the same result as pressing is the first part of an expression it may be omitted. o IT]. If IT] Examples: 1. 5 x (8 -;- 2) = 20 Press ~ 0 IT] [!] EJ rn rn G (4 x (5 + 6» = 144 Press 12 [1] 3 IT] 4 0 IT] 5 [±] 6 rn rn G (12 x (3 + 8)2) = 8730 Press 18 [±] 6 IT] 12 0 IT] 3 [±] 8 rn [EJ 2 rng x «(23 + 6) x 3) ;- 3) = 3984 Press 48 IT] IT] IT] 23 [±] 6 rn 0 3 rn El 12 El 3 rn El Exercise 0.4 Calculate (using a calculator): 1. (2,4) 8 - m ,4 2. 2,6(3 +,1) - M 3. ~~ - -.}9,3 - (3,2)' 6 ~ J(28,4)2 + (6,3)2 - l3~62 + (O,9r 2 17

30 5 (12,3)3 + (2,7) (16,7) 6. (11,2)43 - (4,1~'6 + J1D 7. ~1r (2,W - \!' (l~',~)4 -,r + (2,6)' 9. (3~)4 + (11,2)41 10 (22 W1 + (4,7) 3 - J12 6 _ rr., 3,2 ' (e - 14,6) 12. (25,1-3,8)(4,2 - n) 13. «2,1)3-4,1)e x (18 + 2) (8 + (11-4) 3) Answers to Exercise , , , , , , , , , , , , ,

31 Module 1 Factors and fractions 1.1 The common factor Objectives and overview On completion of this section, you should be able to factorise polynomials containing common factors General The process where a number, term or expression is written as a product is called factorisation (or resolving into factors) Common factors This means that the same factor is present in every term. For example: x is a common factor in ax + bx. Example: Resolve ax + bx into factors. Solution: (i) Short method: ax + bx = x(a + b) Here we say that x is "taken out' as a common factor. This is a very handy and often used method, but it is not really a valid mathematical operation, e.g. sin 2A =f:: 2 sin A. The answer can easily be tested by removing the brackets by multiplying. 19

32 (ii) Taking out could be addition: ax + bx = (a + b)x [a of x and b of x is (a + b) of xl It can be better seen in: 2x + 3x = (2 + 3)x (iii) Taking out could be multiplication: Multiplication by the identity element for multiplication namely = ~ = ~ etc. leaves the expression unchanged. For example: x 14 x x 14 x 1 14 :. ax + bx x = -(ax + bx) ~(x x) = 1" a x + b x x = 1"(a.l + b.l) = x(a + b) [x by 1] [ x by ~ into the bracket] Note: When factorising the highest common factor must be 'taken out'. Examples: Solve into factors: (a) 2ax + 8x - 16a D 2 d 2 (b) Solutions: (a) 2ax + 8x - 16a = 2(ax + 4x - 8a) 20

33 or 2ax + 8x - 16a 2 = - (2ax + 8x - 16a) 2 = ~ (2ax + 8x _ 16a) = 2 [ax + 4x - 8a] [ x by ~ into the bracket] (b) 0 2 d = 21(02 "2 + d~ ') or 0 2 d 2 4+"2 = ~(~2 + ~2) = 2 1( '2d 2) [x by 2 into the bracket] = ~(~2 + d 2 ) The alternative method is very handy in the following examples. Examples: Factorsie 1 + x 2 so that one factor must be: 1 (a) - x (b) x (c) x 2 Solutions: (a) [ A factor must be ~ :. x by ~J X = -(1 + x 2) x 1 J = -(x. 1 + x.x-) x 1 3 = -(x + x) x [Divide ~ into.!..x and x by xl x x 21

34 (b) x ) = -(1 + x-, x ( 1 )1) = x l. x+ x-.;: = x(~ + X) (c) 1 + x 2 [ A factor is x... x by ~J IVI de -xold Into -.x an x by -IJ [DO x x x Grouping Sometimes terms have to be rearranged, or grouped, before factors can be found. Example: Resolve ap + ax + 4p + 4x into factors. Solution: ap + ax + 4p + 4x [Group terms 1 and 2 as well as 3 and 4] = a(p + x) + 4(p + x) [a and 4 are common factors. Take out a and 4] = (p + x)(a + 4) [Take out (p + x) from term 1 and term 2] or ap + ax + 4p + 4x ( ap ax) (4 P 4X) =a a-+a- +4 "4+"4 = a(p + x) + 4(p + x) 22

35 = (P + x)(a. p + x + 4. P + x) P+x P+x = (P + x)(a + 4) The 'take out' method is shorter in this case. Example: 6x + 2x x [Group terms 1 and 2 as well as 3 and 4] = 2x(3 + x) + 1(3 + x) [(3 + x) is now also common] = (3 + x)(2x + 1) or 6x + 2x x 6X 2X2) 1(3 x) =2x( 2x+ 2x = 2x(3 + x) + 1 (3 + x) = (3 + x) (2X(3 + x) + 1(3 + x») 3+x 3+x = (3 + x)(2x + 1) [ Xb ~] y 3 + x Grouping can be applied on 4, 6, 8 terms etc. Example: 2a - 3ax + ac + 2p - 3px + pc [Group first 3 and last 3 terms] = a[2-3x + c] + p[2-3x + c] [(2-3x + c is now common] = (2-3x + c)(a + p) or 2a - 3ax + ac + 2p - 3px + pc = a(2a 3a_x + ac) + p(2 P 3p_x + pc) [X by ~ and e] a a a p p pap = a(2-3x + c) + p(2-3x + c) = (2 _ 3x + C)[a(2-3x + c) + p_(_2_-_3x_ _+_C)] 2-3x + c 2-3x + c [ X b 2-3x + c] y 2-3x + c = (2-3x + c)(a + p) 23

36 Exercise 1.1 Resolve the following into prime (smallest) factors: 1. 12x - 4xy 2. 3ab + 4eb 3. 4nD 2 + 4nd 2 nd 2 nd y(x - 2) + e(x - 2) 6. 4(x + 2) + x 2 + 2x 7. ax 2 + bx - a 2 x - ab 8. 3a - 2ax + ax 2 + 3b - 2bx + bx x x ex 2-2ex - 4e 10. 3a(a + 2b + e) - 3b(-a - 2b - c) 11. 2nJe + 6Je 12. a + x(x 2 + X + a) 13. yz - xz + 3(x - y) 14. Write ~ as a factor of x + x 2 X. 1 ~ f Wnte - as a lactor "2 x x x 16. Write x 2 as a factor of (1 + 2x - 3x 2 ) ~ k: J2X 17. v2x + 2av2x - 3a 18. a(x - y) - bx + by 19. x 2 (x + 1) + 12(x + 1) - 7x(x + 1) 20. xy2 - Y + p(l - xy) Answers to Exercise x(3 - y) 2. b(3a + 4e) 3. 4n(D 2 + d 2) 4. ~D2 + d 2 ) 5. (x - 2)(y + e) 6. (x + 2)(4 + x) 7. (ax + b)(x - a) 8. (3-2x + x 2 )(a + b) 9. (x 2-2x - 4)(4 + e) 24

37 10. 3(a + 2b + c)(a + b) 11. 2Je (n + 3) 12. (a + x 2 )(1 + x) 13. (x - y)(3 - z) ;(x 3 + x 4 ) x- IS. ~(X +1 +~) 16. X2(~ + ~ - 3) x 2 X 17. J2X(1+ 2a - 3~) 18. (x - y)(a - b) 19. (x + 1)(x 2-7x + 12) 20. (xy - 1)(y - p) 1.2 The quadratic trinomial Objectives and overview On completion of this section you should be able to: 1. Solve quadratic trinomials with no limitations on any term. 2. Solve quadratic trinomials containing common factors General Generally a quadratic trinomial looks like: ax 2 + bx + c or ax 2 + bxy + cy2 where a, band c are constants. Quadratic trinomials originate when certain word problems are transferred to symbolic language (mathematised) or if binomials are multiplied. For example: (x + 2)(x - 1) = x 2 + X - 2 (3x + 2)(x + 1) = 3x 2 + 5x + 2 (a - x)(2a - 3x) = 2a 2-5ax + 3x 2 (x - 3)(2x + 1) = 2x 2-5x - 3 (2x + a)(3x + 2a) = 6x 2 + tax + 2a 2 25

38 Note that if the sign of the third terms is +, the signs of the factors will be the same. If the sign of the third term is -, the signs of the factors will differ The solvability of trinomials The factors ofax l + bxy + cyl can easily be solved if the product of a and c can be broken up into factors p and q so that the sum is equal to b, p + q = b. It is important to remember that if the sign of the last term is positive, p and q must have the same sign. If the sign of the last term is negative the signs ofp and q will differ..". ax l + bxy + cyl becomes ax 2 + (p + q) xy + cyl. Example: x 2 _ x - 6 = x 2 + (p + q)x - 6 = Xl + (-3 + 2)x - 6 = Xl - 3x + 2x - 6 = x(: - ~) + 2(2; _~) = x(x - 3) + 2(x - 3) = (x - 3)[X(X - 3) + 2(x - 3)J x-3 x-3 = (x - 3)(x + 2).".a = 1, b = - 1 and c = - 6.".ac=-6 Factors of 6 1 and 6 cannot result in an 2 can result in -1 if p = -3 and q = 2.".p+q=-1 Example: x 2 _ 5x + 6 = Xl + (p + q)x + 6 = x 2 + (-3-2)x + 6 = x 2-3x - 2x + 6 = x(x 2 _ 3X) _ 2(-2X + ~) x x -2-2 = x(x - 3) - 2(x - 3) = (x _ 3)[X(X - 3) _ 2(x - 3)J x-3 x-3 = (x - 3)(x - 2) a = 1, b = - 5 and c = 6 ac = 6 1 x 6 => = -5.". P = - 6 and q = 1 but the sign of 6 must be +... the signs of p and q must be the same 2 x 3 => -3-2 = - 5.". P = - 3 and q =

39 Ifp = - 6 and q = 1 it is not possible to find factors. x 2 _ 5x + 6 = x 2 + (p + q)x + 6 = x 2 + (-6 + l)x + 6 = X 2 (;2 6x :x)x + 6 = x x - x + l(x + 6) = x(x - 6) + l(x + 6) Example: 6x 2-9x - 42 = 6x 2 + (p + q)x - 42 = 6x 2 + ( )x - 42 = 6x 2-21x + 12x - 42 = 3X(6X 2 _ 21X) + 6(12X _ 42) 3x 3x 6 6 = 3x(2x - 7) + 6(2x - 7) = (2x _ 7)[3X(2X - 7) + 6(2x - 7)J 2x-7 2x-7 = (2x - 7)(3x + 6) = (3x + 6)(2x - 7) 3X 6) ( = 3 3 +"3 (2x - 7) = 3(x + 6)(2x - 7) or 6x 2-9x - 42 = 3(6X 2 _ 9x _ 42) = 3(2x 2-3x - 14) = 3[2x 2 + (p + q)x - 14] = 3[2x 2 + (-7 + 4)x - 14] = 3[2x 2-7x + 4x - 14] = 3[x(2;2 - ~) + 2(~ )J a = 6, b = - 9 and c = -42 ac = 6( -42) = x x x 84 4 x 63 6 x 42 7 x 36 9 x x 21 ~ = p = - 21 and q = 12 The signs ofp and q must differ.... a = 2, b = - 3 and c = -14 ac = 2(-14) = x 28 2 x 14 4x7~-7+4=-3... p = - 7 and q = 4 27

40 = 3[x(2x - 7) + 2(2x - 7)] = 3{ (2X _ 7)[X(2X - 7) + 2(2x - 7)J 2x - 7 2x - 7 = 3{ (2x - 7)(x + 2)} = 3(2x - 7)(x + 2) Example: 2x 2-5xy + 3y 2 = 2x 2 + (p + q)xy + 3y 2 = 2x 2 + (-3-2)xy + 3y 2 = 2x 2-3xy - 2xy + 3y 2 6 a = 2, b = - 5 and c = 3 ac=2x3=6 = x(2x 2 _ 3XY) _ y(-2xy + 3 y2 ) : X_IS ~ -6 + I x x -y-y But the signs of p and q = x(2x - 3y) - y(2x - 3y) must not differ = (2x _ 3 y )[X(2X --- 3y) _ y(2x - 3 y )J 3 x 2 ~ d 2 = x - 3y 2x - 3y.'. p = - an q = - = (2x - 3y)(x - y) Example: 10-3x - x (p + q)x - x (-5 + 2)x - x x + 2x - x 2 = scso - s;) + x( ~ -:) = 5(2 - x) + x(2 - x) a = - 1, b = - 3 and c = 10 ac = x 10 2 x 5 ~ = -3.'. p = - 5 and q = 2 = (2 _ X)[5(2 - x) + x(2 - X)J 2-x 2-x = (2 - x)(5 + x) Experienced people can write it out in a much shorter form The traditional method You must also know that if the sign of the last term is negative then the signs of the factors must differ and when the sign of the last term is positive the signs of the factors must be the same. Resolving involves guessing the factors in the brackets. 28

41 Example: x 2-2x - 3 = ( )( ) Method: 1. Write down the factors ofthe first term underneath each other. 2. Write down the factors of the last term underneath each other just to the right of x x 3. Multiply across each other and write down the answers on the same horizontal plane as the arrows. Just to the right ofi 4. Try to find the middle term by adding or subtracting 3x and x. Remember that ifthe sign ofthe last term is -, the signs of the factors must differ. If the sign of the last term is +, the signs of the factors must be the same + or The signs of the answers are displaced to the arrow points on the same horizontal plane. x x x 3 x XX 3 3x x 1 x X: -3x +x -2x ~ x - 3-3x x+~+x -2x :. x 2 _ 2x - 3 = (x - 3)(x + 1) As a test, the factors can be multiplied to result in the original expression. Exercise 1.2 Resolve into factors: 1. x 2 + X x 2-5x x 2 + 8x x x p 2-21p a 2-85a a 2 + 5ab + 3b 2 2. x 2 + 9x x x x x x 2-26x a a a 2 + 2ab + b x 2 + xy - 2y 2 29