Chapter 2 PDF - Algebraic Foundations

“c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 55 — #1 2 Algebraic foundations LEARNING SEQUENCE 2.1 Overview....................................................................................................................................................................56 2.2 Algebraic skills........................................................................................................................................................ 57 2.3 Pascal’s triangle and binomial expansions.................................................................................................. 67 2.4 The binomial theorem...........................................................................................................................................71 2.5 Sets of real numbers.............................................................................................................................................80 2.6 Surds...........................................................................................................................................................................87 2.7 Review........................................................................................................................................................................ 99 Fully worked solutions for this topic are available online. “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 56 — #2 2.1 Overview Hey students! Bring these pages to life online Watch Engage with Answer questions videos interactivities and check results Find all this and MORE in jacPLUS 2.1.1 Introduction In 2017 the Cassini spacecraft disintegrated in Saturn’s atmosphere, having successfully completed a 7-year mission to observe that planet and its moons. In all, it completed 22 orbits of Saturn, sending extraordinary images and information back to Earth. In an earlier mission, the Voyager 1 spacecraft passed Pluto in 1990, and in 2004 it left our solar system. On board Voyager 1 was a time capsule carrying information about Earth and our achievements for, should they exist, any intelligent alien life forms it may reach. None of this would have been possible without mathematics. Mathematics and physics are essential to the launch and success of all space missions. It has been argued that should there ever be communication between Earthlings and intelligent extra-terrestrials, communication will be through mathematics. This is because mathematics is universal. It is universally true that Pythagoras’ theorem, a2 + b2 = c2 , holds in every country on Earth and, by extension, in any galaxy in the universe. Expressing the rule for Pythagoras’ theorem in symbols gives an example of the succinct nature of algebra. It is the language of mathematics — the unifying thread that has evolved through the ages — that underlies its different branches. Elementary algebra, as studied at school, seeks to establish the fundamentals vital for confident and automatic use of this language. There are other algebras, higher-order ones such as group theory, rings and fields, that have developed only since the 19th century. Despite its abstract nature, research in these fields is highly valued today for its applications in cryptography, encryption and other aspects of internet security. Two modern-day mathematicians include Cheryl Praeger, an Australian who works in group theory and combinatorics, and the late Maryam Mirzakhani, so far the only female recipient of the Fields medal — the highest award in Mathematics — for her work in the dynamics and geometry of complex surfaces. KEY CONCEPTS This topic covers the following key concepts from the VCE Mathematics Study Design: use of symbolic notation to develop algebraic expressions and represent functions, relations, equations, and systems of simultaneous equations substitution into, and manipulation of, these expressions recognition of equivalent expressions and simplification of algebraic expressions involving different forms of polynomial and power functions, the use of distributive and exponent laws applied to these functions, and manipulation from one form of expression to an equivalent form. Note: Concepts shown in grey are covered in other topics. Source: VCE Mathematics Study Design (2023–2027) extracts © VCAA; reproduced by permission. 56 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 57 — #3 2.2 Algebraic skills LEARNING INTENTION At the end of this subtopic you should be able to: expand and factorise linear and simple quadratic expressions with integer coefficients by hand factorise the sum or difference of two cubes simplify algebraic fractions. This topic covers some of the algebraic skills required for the foundation to learning and understanding of Mathematical Methods. Some basic algebraic techniques will be revised and some new techniques will be introduced. 2.2.1 Review of expansion and factorisation Expansion The distributive law is fundamental in expanding to remove brackets. Distributive law a(b + c) = ab + ac Some simple expansions include: (a + b)(c + d) = ac + ad + bc + bd (a + b)2 = a2 + 2ab + b2 (a − b)2 = a2 − 2ab + b2 (a + b)(a − b) = a2 − b2 WORKED EXAMPLE 1 Expanding and simplifying an expression Expand 2(4x − 3)2 − (x − 2)(x + 2) + (x + 5)(2x − 1) and state the coefficient of the x term. THINK WRITE 1. Expand each pair of brackets. 2(4x − 3)2 − (x − 2)(x + 2) + (x + 5)(2x − 1) Note: The first term contains a perfect square, = 2(16x2 − 24x + 9) − (x2 − 4) + (2x2 − x + 10x − 5) the second a difference of two squares and the third a quadratic trinomial. 2. Expand fully, taking care with signs. = 32x2 − 48x + 18 − x2 + 4 + 2x2 + 9x − 5 3. Collect like terms together. = 33x2 − 39x + 17 4. State the answer. The expansion gives 33x2 − 39x + 17 and the Note: Read the question again to ensure the coefficient of x is −39. answer given is as requested. TOPIC 2 Algebraic foundations 57 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 58 — #4 TI | THINK DISPLAY/WRITE CASIO | THINK DISPLAY/WRITE 1. On a Calculator page, 1. On a Main screen, press MENU and select: complete the 3. Algebra entry line as: 3. Expand expand Complete the (2 × (4x − 3)2 − (x − 2) entry line as: (x + 2) + (x + 5)(2x − 1)) expand Then press EXE. (2 × (4x − 3)2 − (x − 2) (x + 2) + (x + 5)(2x − 1)) Then press ENTER. 2. The answer appears on The expansion gives 2. The answer appears on The expansion gives the screen. 33x2 − 39x + 17. the screen. 33x2 − 39x + 17. Factorisation Some simple factors include: Factorised form common factors (a + b)(c + d) the difference of two perfect squares perfect squares and factors of other quadratic trinomials. to rise exp is equal to and fa c ac + ad + bc + bd Expanded form A systematic approach to factorising is displayed in the following diagram. Common factor? How many terms? Two terms Three terms Four or more terms Difference of two squares? Quadratic trinomial? Grouping? a2 – b2 = (a + b)(a – b) Perfect squares? ac + ad + bc + bd a2 + 2ab + b2 = (a + b)2 = a(c + d) + b(c + d) a2 – 2ab + b2 = (a – b)2 = (a + b)(c + d) 58 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 59 — #5 Grouping terms is commonly referred to as grouping ‘2 and 2’ and grouping ‘3 and 1’, depending on the number of terms grouped together. For example, as the first three terms of a2 + 2ab + b2 − c2 are a perfect square, grouping ‘3 and 1’ would create a difference of two squares expression, allowing the whole expression to be factorised. a2 + 2ab + b2 − c2 = (a2 + 2ab + b2 ) − c2 = (a + b)2 − c2 = (a + b − c)(a + b + c) WORKED EXAMPLE 2 Factorising algebraic expressions Factorise: a. 3x2 − 12y2 b. x2 + 7x − 8 c. 3x2 − 17x + 10 d. 2x3 + 5x2 y − 12y2 x e. 4y2 − x2 + 10x − 25 f. 7(x + 1) − 8(x + 1) + 1 using the substitution a = (x + 1). 2 THINK WRITE ( ) a. 1. Take out the common factor. a. 3x2 − 12y2 = 3 x2 − 4y2 ( ) 2. Recognise the difference of two squares. = 3 x2 − (2y)2 3. Factorise using the difference of two = 3 (x − 2y) (x + 2y) squares, a2 − b2 = (a − b) (a + b). b. 1. Recognise the trinomial ax2 + bx + c. b. x2 + 7x − 8, where a = 1, b = 7 and c = −8 2. Find two numbers that multiply to the ac = −8, b = 7 outside product, ac, and add to the The numbers are −1 and 8. middle, b. 3. Split the middle term of the given equation. x2 + 7x − 8 = x2 − 1x + 8x − 8 4. Pair and factorise. = x2 − 1x + 8x − 8 = x (x − 1) + 8 (x − 1) 5. Take out the common factor. = (x − 1) (x + 8) Note: The middle term may be split as 7x = −1x + 8x or 7x = 8x − 1x. c. 1. Recognise the trinomial ax2 + bx + c. c. 3x2 − 17x + 10, where a = 3, b = −17 and c = 10 2. Find two numbers that multiply to the ac = 30, b = −17 outside product, ac, and add to the The numbers are –15 and –2. middle, b. 3. Split the middle term of the given equation. 3x2 − 17x + 10 = 3x2 − 2x − 15x + 10 4. Pair and factorise. = 3x2 − 2x − 15x + 10 = x (3x − 2) − 5 (3x − 2) 5. Take out the common factor. = (3x − 2) (x − 5) ( ) d. 1. Take out the common factor. d. 2x3 + 5x2 y − 12xy2 = x 2x2 + 5xy − 12y2 2. To factorise the expression in the brackets, The numbers are 8 and −3. find two numbers that multiply to the outside product (2x − 12 = −24) and add to the middle (−3 + 8 = 5). TOPIC 2 Algebraic foundations 59 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 60 — #6 ( ) ( ) 3. Split the middle term of the expression in x 2x2 + 5xy − 12y2 = x 2x2 − 3xy + 8xy − 12y2 brackets. ( ) 4. Pair and factorise. = x 2x2 − 3xy + 8xy − 12y2 = x (x (2x − 3y) + 4y (2x − 3y)) 5. Take out the common factor and write the = x (2x − 3y) (x + 4y) three factors. e. 1. The last three terms of the expression can be e. 4y2 − x2 + 10x − 25 = 4y2 − (x2 − 10x + 25) grouped together to form a perfect square. 2. Use the grouping ‘3 and 1’ technique to = 4y2 − (x − 5)2 create a difference of two squares. = (2y)2 − (x − 5)2 3. Factorise the difference of two squares. = [2y − (x − 5)][2y + (x − 5)] 4. Remove the inner brackets to obtain the = (2y − x + 5)(2y + x − 5) answer. f. 1. Substitute a = (x + 1) to form a quadratic f. 7(x + 1)2 − 8(x + 1) + 1 trinomial in a. = 7a2 − 8a + 1 where a = (x + 1) 2. Factorise the trinomial using any suitable = (7a − 1)(a − 1) method. 3. Substitute (x + 1) back in place of a. = (7(x + 1) − 1)((x + 1) − 1) 4. Remove the inner brackets and simplify to = (7x + 7 − 1)(x + 1 − 1) obtain the answer. = (7x + 6)(x) = x(7x + 6) TI | THINK DISPLAY/WRITE CASIO | THINK DISPLAY/WRITE b. 1. On a Calculator page, press b. 1. On a Main screen, complete MENU and select: the entry line as: 3. Algebra factor (4y2 − x2 + 10x − 25) 2. Factor Then press EXE. Complete the entry line as: factor (4y2 − x2 + 10x − 25) Then press ENTER. 2. The answer appears on the −(x − 2y − 5)(x + 2y − 5) 2. The answer appears on the −(x + 2y − 5) screen. screen. (x − 2y − 5) 2.2.2 Factorising sums and differences of two perfect cubes Check the following by hand or by using CAS technology. Expanding (a + b)(a2 − ab + b2 ) gives a3 + b3 , the sum of two cubes. Expanding (a − b)(a2 + ab + b2 ) gives a3 − b3 , the difference of two cubes. 60 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 61 — #7 Hence, the factors of the sum and difference of two cubes are as follows. Sum and difference of two perfect cubes a3 + b3 = (a + b)(a2 − ab + b2 ) a3 − b3 = (a − b)(a2 + ab + b2 ) WORKED EXAMPLE 3 Factorising perfect cubes Factorise: a. x3 − 27 b. 2x3 + 16. THINK WRITE 3 a. 1. Express x − 27 as a difference of two cubes. a. x3 − 27 = x3 − 33 2. Apply the factorisation rule for the difference Using a3 − b3 = (a − b)(a2 + ab + b2 ) with of two cubes. a = x, b = 3, x3 − 33 = (x − 3)(x2 + 3x + 32 ) 3. State the answer. ∴ x3 − 27 = (x − 3)(x2 + 3x + 9) b. 1. Take out the common factor. b. 2x3 + 16 = 2(x3 + 8) 2. Express x3 + 8 as a sum of two cubes. = 2(x3 + 23 ) 3. Apply the factorisation rule for the sum of Using a3 + b3 = (a + b)(a2 − ab + b2 ) with two cubes. a = x, b = 2, x3 + 23 = (x + 2)(x2 − 2x + 22 ) 2(x3 + 23 ) = 2(x + 2)(x2 − 2x + 22 ) 4. State the answer. ∴ 2x3 + 16 = 2(x + 2)(x2 − 2x + 4) 2.2.3 Algebraic fractions The same methods used to simplify, add, subtract, multiply or divide arithmetic fractions are used to simplify algebraic fractions. Simplifying algebraic fractions An algebraic fraction can be simplified by cancelling any common factor between its numerator and its denominator. For example: ab + ac a(b + c) = ad ad b+c = d Multiplication and division of algebraic fractions For the product of algebraic fractions, once numerators and denominators have been factorised, any common factors can then be cancelled. The remaining numerator terms are usually left in factors, as are any remaining denominator terms. For example: + c) × d a(b (a + c) (b + c)(a + c) = ad b b TOPIC 2 Algebraic foundations 61 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 62 — #8 Note that b is not a common factor of the numerator so it cannot be cancelled with the b in the denominator. As in arithmetic, to divide by an algebraic fraction, multiply by its reciprocal. a c a d ÷ = × b d b c WORKED EXAMPLE 4 Simplifying algebraic fractions Simplify: x2 − 2x x4 − 1 1 + x2 a. b. ÷ x2 − 5x + 6 x−3 3−x THINK WRITE x2 − 2x x(x − 2) a. 1. Factorise both the numerator and the a. 2 = denominator. x − 5x + 6 (x − 3)(x − 2) Note: The numerator has a common factor; the denominator is a quadratic trinomial. x − (x 2) 2. Cancel the common factor in the numerator = and denominator. (x − 3) − (x 2) x 3. Write the fraction in its simplest form. = x−3 No further cancellation is possible. x4 − 1 1 + x2 x4 − 1 3 − x b. 1. Change the division into multiplication by b. ÷ = × replacing the divisor by its reciprocal. x−3 3−x x − 3 1 + x2 2 2. Factorise where possible. Since x4 − 1 = (x2 ) − 12 Note: The aim is to create common factors of = (x2 − 1)(x2 + 1) both the numerator and denominator. For this then: reason, write (3 − x) as −(x − 3). x4 − 1 3 − x × x − 3 1 + x2 (x2 − 1)(x2 + 1) −(x − 3) = × x−3 1 + x2 2 2 (x − 1)(x + 1) − − (x 3) = × x−3 1+x 2 (x2 − 1) −1 3. Cancel the two sets of common factors of the = × numerator and denominator. 1 1 −(x2 − 1) 4. Multiply the remaining terms in the = numerator together and the remaining terms 1 in the denominator together. 5. State the answer. = −(x2 − 1) = 1 − x2 62 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 63 — #9 TI | THINK DISPLAY/WRITE CASIO | THINK DISPLAY/WRITE b. 1. On a Calculator page, b. 1. On a Main screen, complete complete the entry line as: the entry line as: x4 − 1 1 + x2 x4 − 1 1 + x2 / / x−3 3−x x−3 3−x Then press ENTER. Press EXE, then press SIMP to simplify the answer. 2. The answer appears on the −(x2 − 1) 2. The answer appears on the −x2 + 1 screen. screen. Addition and subtraction of algebraic fractions Factorisation and expansion techniques are often required when adding or subtracting algebraic fractions. Denominators should be factorised in order to select the lowest common denominator. Express each fraction with this lowest common denominator. Simplify by expanding the terms in the numerator and collect any like terms together. WORKED EXAMPLE 5 Adding and subtracting algebraic fractions 2 1 x Simplify − +. 3x + 3 x − 2 x2 − x − 2 THINK WRITE 2 1 x 1. Factorise each denominator. − + 3x + 3 x − 2 x2 − x − 2 2 1 x = − + 3(x + 1) (x − 2) (x + 1)(x − 2) 2 × (x − 2) 1 × 3(x + 1) x×3 2. Select the lowest common denominator = − + and express each fraction with this as its 3(x + 1)(x − 2) 3(x + 1)(x − 2) 3(x + 1)(x − 2) denominator. 2(x − 2) − 3(x + 1) + 3x 3. Combine the fractions into one fraction. = 3(x + 1)(x − 2) 2x − 4 − 3x − 3 + 3x 4. Expand the terms in the numerator. = Note: It is not necessary to expand the 3(x + 1)(x − 2) denominator terms. 2x − 7 5. Collect like terms in the numerator and state = the answer. 3(x + 1)(x − 2) Note: Since there are no common factors between the numerator and the denominator, the fraction is in its simplest form. TOPIC 2 Algebraic foundations 63 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 64 — #10 2.2 Exercise Students, these questions are even better in jacPLUS Receive immediate Access Track your feedback and access additional results and sample responses questions progress Find all this and MORE in jacPLUS Technology free 1. Expand the following and simplify where appropriate. a. 4m(m − 2) + 3m b. 5(m2 − 3m + 2) − (m + 2) c. (x − 3)(x + 5) 2 d. (3m − 2)(5m − 4) e. (4 − 3x)(4 + 3x) f. (2x − 5) 2. Expand and simplify the following. a. 2(x − 5)(x + 5) − 3(2x − 3) b. 3 − 2(x + 5)(3x − 2) 2 c. (3 − 2x)(x − 5) − (x + 3)(x + 4) d. 3(2x − 1)(2x + 1) + (x − 5) 3. WE1 Expand 3(2x + 1)2 + (7x + 11)(7x − 11) − (3x + 4)(2x − 1) and state the coefficient of the x term. 4. Expand each of the following expressions. 2 a. (2x + 3) b. 4a(b − 3a)(b + 3a) c. 10 − (c + 2)(4c − 5) 2 3 3 3 d. (5 − 7y) e. (3m + 4n)(3m − 4n) f. (x + 1) 5. Expand and simplify the following, and state the coefficient of the x term. a. 2(2x − 3)(x − 2) + (x + 5)(2x − 1) b. (2 + 3x)(4 − 6x − 5x2 ) − (x − 6)(x + 6) 2 c. (4x + 7)(4x − 7)(1 − x) d. (x + 1 − 2y)(x + 1 + 2y) + (x − 1) e. (3 − 2x)(2x + 9) − 3(5x − 1)(4 − x) f. x2 + x − 4(x2 + x − 4) 6. Factorise the following. a. x2 − 36 b. 4 − 25a2 c. 9m2 − 1 2 d. 4a2 − 64 e. 2m2 − 98x2 f. 1 − 9(1 − m) 7. Factorise the following. a. x2 − 9x + 18 b. x2 − 6x + 9 c. x2 + 7x − 60 d. 4x2 + 4x − 15 e. 4x2 − 20x + 25 f. 8x2 − 48xy + 72y2 8. WE2 Factorise: a. 5x − 45y2 2 b. x2 − 9x − 10 c. 8x2 − 14x − 15 d. 4x3 − 8x2 y − 12xy2 e. 9y2 − x2 − 8x − 16 f. 4 (x − 3)2 − 3 (x − 3) − 22 using the substitution a = x − 3. 9. Factorise: 2 a. 49 − 168x + 144x2 b. 2(x − 1) + 13(x − 1) + 20 2 c. 40(x + 2) − 18(x + 2) − 7 d. 144x2 − 36y2 e. 3a2 x + 9ax − a − 3 f. 16x2 + 8x + 1 − y2 64 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 65 — #11 10. Fully factorise the following. a. x3 + 2x2 − 25x − 50 b. 100p3 − 81pq2 c. 4n2 + 4n + 1 − 4p2 d. 49(m + 2n)2 − 81(2m − n)2 e. 13(a − 1) + 52(1 − a)3 f. a2 − b2 − a + b + (a + b − 1)2 11. Use a substitution method to factorise the following. 2 2 a. (x + 5) + (x + 5) − 56 b. 2(x + 3) − 7(x + 3) − 9 2 c. 70(x + y) − y(x + y) − 6y2 d. x4 − 8x2 − 9 ( )2 ( ) 2 2 2 2 2 1 2 1 e. 9(p − q) + 12(p − q ) + 4(p + q) f. a a+ − 4a a+ + 4a2 a a 12. WE3 Factorise: a. x3 − 125 b. 3 + 3x3 13. Factorise the following. a. x3 − 8 b. x3 + 1000 c. 1 − x3 3 3 4 3 d. 27x + 64y e. x − 125x f. (x − 1) + 216 14. Fully factorise the following. a. xy3 − 27x b. −x3 − 216 c. 3 − 81x3 d. 32x3 + 4m3 e. 27m3 + 64n3 f. 250x3 − 128m3 15. Fully factorise the following. 3 3 a. 24x3 − 81y3 b. 8x4 y4 + xy c. 125(x + 2) + 64(x − 5) 3 3 5 3 2 2 3 5 6 6 d. 2(x − y) − 54(2x + y) e. a − a b + a b − b f. x − y 16. WE4 Simplify: 2 x + 4x x4 − 64 x2 + 8 a. b. ÷ x2 + 2x − 8 5−x x−5 17. Simplify the following algebraic fractions. x x−2 5 10 a. × b. ÷ (x + 1)(x − 2) 3x x(3x + 1) x(x + 3) 2 x + 5x + 6 16 − 9x2 2x + 10 c. d. × 4x + 8 8 + 6x 3x − 4 4x 18x2 − 6x 2x2 − 3x − 5 3x2 − 5x − 12 e. ÷ f. × 3x2 + 5x + 2 9x2 − 1 2x2 − 11x + 15 3x2 + 7x + 4 18. Simplify the following. 3x2 − 7x − 20 x3 + 4x2 − 9x − 36 a. b. 25 − 9x2 x2 + x − 12 (x + h)3 − x3 2x2 1 − 9x2 c. d. 3 2 × 2 h 9x + 3x 18x − 12x + 2 m − 2m2 n 3 m2 − 4n2 1−x 3 1 − x2 1 + x + x2 e. ÷ f. × ÷ m3 + n3 m2 + 3mn + 2n2 1 + x3 1 + x2 1 − x + x2 6 1 2x 19. WE5 Simplify + − 2. 5x − 25 x − 1 x − 6x + 5 TOPIC 2 Algebraic foundations 65 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 66 — #12 20. Express each of the following as a single algebraic fraction. 2x 5x x 3x 4 5 a. + b. − c. + 3 4 7 2 x−3 x+5 x 5 x−3 x+2 3 1 5 d. − e. − f. 2 − + 3x − 1 1 − 2x 2x + 1 x − 1 x −9 x+3 x−3 Technology active 21. Simplify the following expressions. 4 4 4 3 5 a. + b. − + x + 1 x − x2 2 2 x −4 x+2 x−2 5 4 3 1 2 1 c. + + 2 d. 2 + 2 − 2 x + 6 5 − x x + x − 30 4y − 36y + 81 4y − 81 2y − 9y 22. a. Expand (2 + 3x)(x + 6)(3x − 2)(6 − x). b. Factorise x2 − 6x + 9 − xy + 3y. 3 c. Factorise 2y4 + 2y(x − y). 23. Expand the following. 2 a. (g + 12 + h) 2 b. (2p + 7q) (7q − 2p) c. (x + 10)(5 + 2x)(10 − x)(2x − 5) 24. Simplify the following. ( ) x3 − 125 5 4 3 16x2 − 1 a. × 3 b. − ÷ 2 x − 25 x + 5x2 + 25x 2 x + 1 (x + 1) 2 x + 2x + 1 ( ) 1 p q3 7 5 c. − 2 − d. (a + 6b) ÷ − p − q p − q2 p4 − q4 a2 − 3ab + 2b2 a2 − ab − 2b2 25. Using CAS technology: a. expand (x + 5)(2 − x)(3x + 7) 3 3 b. factorise 27(x − 2) + 64(x + 2) 3 8 c. simplify +. x−1 x+8 2.2 Exam questions Question 1 (1 mark) TECH-ACTIVE MC The coefficient of x in the expansion of x (x − 2) (x + 1) + (x − 3)2 is A. 12 B. 8 C. 4 D. −4 E. −8 Question 2 (1 mark) TECH-ACTIVE MC The factors of 9 (2x − 1)2 − 9 are ( ) 1 A. (1, 0) B. 2x − 1, 9 C. − ,9 D. 36x, x − 1 E. 36x2 − 36x 2 Question 3 (1 mark) TECH-FREE Factorise 8 (x + 3)2 + 24 (x + 3) + 16. More exam questions are available online. 66 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 67 — #13 2.3 Pascal’s triangle and binomial expansions LEARNING INTENTION At the end of this subtopic you should be able to: expand and simplify perfect cubes use binomial coefficients from Pascal’s triangle to expand higher powers of (a + b). 2.3.1 Expansions of perfect cubes The perfect square (a + b)2 may be expanded quickly by the rule (a + b)2 = a2 + 2ab + b2. The perfect cube (a + b)3 can also be expanded by a rule. This rule is derived by expressing (a + b)3 as the product of repeated factors and expanding. (a + b)3 = (a + b)(a + b)(a + b) = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2 ) = a3 + 2a2 b + ab2 + ba2 + 2ab2 + b3 = a3 + 3a2 b + 3ab2 + b3 Therefore, the rules for expanding a perfect cube are as follows. Perfect cubes (a + b)3 = a3 + 3a2 b + 3ab2 + b3 (a − b)3 = a3 − 3a2 b + 3ab2 − b3 Features of the rule for expanding perfect cubes The powers of the first term, a, decrease as the powers of the second term, b, increase. The coefficients of each term in the expansion of (a + b)3 are 1, 3, 3, 1. The coefficients of each term in the expansion of (a − b)3 are 1, −3, 3, −1. The signs alternate − + − in the expansion of (a − b)3. WORKED EXAMPLE 6 Expanding a perfect cube Expand (2x − 5)3. THINK WRITE 1. Use the rule for expanding a perfect cube. (2x − 5)3 Using (a − b)3 = a3 − 3a2 b + 3ab2 − b3 , let 2x = a and 5 = b. (2x − 5)3 = (2x)3 − 3(2x)2 (5) + 3(2x)(5)2 − (5)3 2. Simplify each term. = 8x3 − 3 × 4x2 × 5 + 3 × 2x × 25 − 125 = 8x3 − 60x2 + 150x − 125 3. State the answer. ∴ (2x − 5)3 = 8x3 − 60x2 + 150x − 125 TOPIC 2 Algebraic foundations 67 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 68 — #14 2.3.2 Pascal’s triangle Pascal’s triangle contains many fascinating patterns. Each row from row 1 onwards begins and ends with ‘1’. Each other number along a row is formed by adding the two terms to its left and right from the preceding row. Row 0 1 Row 1 1 1 Row 2 1 2 1 Row 3 1 3 3 1 Row 4 1 4 6 4 1 The numbers in each row are called binomial coefficients. The numbers 1, 2, 1 in row 2 are the coefficients of the terms in the expansion of (a + b)2. (a + b)2 = 1a2 + 2ab + 1b2 The numbers 1, 3, 3, 1 in row 3 are the coefficients of the terms in the expansion of (a + b)3. (a + b)3 = 1a3 + 3a2 b + 3ab2 + 1b3 Each row of Pascal’s triangle contains the coefficients in the expansion of a power of (a + b). To expand (a + b)4 we would use the binomial coefficients 1, 4, 6, 4, 1 from row 4 to obtain: (a + b)4 = 1a4 + 4a3 b + 6a2 b2 + 4ab3 + 1b4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 Notice that the powers of a decrease by 1 as the powers of b increase by 1, with the sum of the powers of a and b always totalling 4 for each term in the expansion of (a + b)4. For the expansion of (a − b)4 the signs would alternate: (a − b)4 = a4 − 4a3 b + 6a2 b2 − 4ab3 + b4 By extending Pascal’s triangle, higher powers of such binomial expressions can be expanded. WORKED EXAMPLE 7 Using Pascal’s triangle in binomial expansions Form the rule for the expansion of (a − b)5 and hence expand (2x − 1)5. THINK WRITE 1. Choose the row in Pascal’s triangle For (a − b)5 , the power of the binomial is 5. Therefore, the that contains the required binomial binomial coefficients are in row 5. The binomial coefficients coefficients. are: 1, 5, 10, 10, 5, 1. 2. Write down the required binomial Alternate the signs: expansion. (a − b)5 = a5 − 5a4 b + 10a3 b2 − 10a2 b3 + 5ab4 − b5 68 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 69 — #15 3. State the values to substitute in To expand (2x − 1)5 , a = 2x, b = 1. place of a and b. 4. Write down the expansion. (2x)5 − 5(2x)4 (1) + 10(2x)3 (1)2 − 10(2x)2 (1)3 + 5(2x)(1)4 − (1)5 5. Evaluate the coefficients and state = 32x5 − 5 × 16x4 + 10 × 8x3 − 10 × 4x2 + 10x − 1 the answer. = 32x5 − 80x4 + 80x3 − 40x2 + 10x − 1 ∴ (2x − 1)5 = 32x5 − 80x4 + 80x3 − 40x2 + 10x − 1 Resources Resourceseses Interactivity Pascal’s triangle and binomial coefficients (int-2554) 2.3 Exercise Students, these questions are even better in jacPLUS Receive immediate Access Track your feedback and access additional results and sample responses questions progress Find all this and MORE in jacPLUS Technology free 3 1. WE6 Expand (3x − 2). 2. Expand the following and simplify where appropriate. 3 3 3 a. (x − 3) b. (2x − 1) c. (x + 4) 3. Expand the following. 3 3 3 a. (3x + 1) b. (1 − 2x) c. (5x + 2y) 4. MC Select the correct statement(s). 3 3 3 A. (x + 2) = x3 + 6x2 + 12x + 8 B. (x + 2) = x3 + 23 C. (x + 2) = (x + 2)(x2 − 2x + 4) 3 3 D. (x + 2) = (x + 2)(x2 + 2x + 4) E. (x + 2) = x3 + 3x2 + 3x + 8 ( )3 5. Expand a + 2b2 and give the coefficient of a2 b2. 6. Copy and complete the following table by making use of Pascal’s triangle. Binomial power Expansion Number of terms Sum of indices in the expansion in each term (x + a)2 (x + a)3 (x + a)4 (x + a)5 7. WE7 Form the rule for the expansion of (a − b)6 and hence expand (2x − 1)6. 4 8. Expand (3x + 2y). TOPIC 2 Algebraic foundations 69 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 70 — #16 Technology active 9. Expand the following using the binomial coefficients from Pascal’s triangle. 5 5 5 a. (x + 4) b. (x − 4) c. (xy + 2) 10. Expand the following using the binomial coefficients from Pascal’s triangle. 4 4 a. (3x − 5y) b. (3 − x2 ) 6 6 11. Expand and simplify (1 + x) − (1 − x) using the binomial coefficients from Pascal’s triangle. 12. Find the coefficient of x2 in the following expressions. 3 2 3 a. (x + 1) − 3x(x + 2) b. 3x2 (x + 5)(x − 5) + 4(5x − 3) 3 3 2 3 c. (x − 1)(x + 2)(x − 3) − (x − 1) d. (2x2 − 3) + 2(4 − x ) 13. Expand and simplify [(x − 1) + y]4. ( )6 x 2 14. Find the term independent of x in the expansion of +. 2 x 4 15. If the coefficient of x2 y2 in the expansion of (x + ay) is 3 times the coefficient of x2 y3 in the expansion of 4 (ax2 − y) , find the value of a. 5 16. Find the coefficient of x in the expansion of (1 + 2x)(1 − x). 4 17. a. Expand (1 + x). b. Using a suitably chosen value for x, evaluate 1.14 using the expansion in part a. 5 4 5 4 4 18. Expand (x + 1) − (x + 1) and hence show that (x + 1) − (x + 1) = x(x + 1). n+1 19. Prove (x + 1) − (x + 1)n = x(x + 1)n. 20. A section of Pascal’s triangle is shown. Determine the values of a, b and c. 45 a b 165 330 220 c 2.3 Exam questions Question 1 (1 mark) TECH-ACTIVE MC The coefficient of x3 in the expansion of (x + 2)5 is A. 120 B. 40 C. 1 D. 8 E. 20 Question 2 (1 mark) TECH-ACTIVE ( )4 2 MC The coefficient of the term independent of in the expansion of x− is x A. 4 B. −6 C. 6 D. −24 E. 24 Question 3 (1 mark) TECH-FREE Expand (x − 2) − (x − 2)3 and hence show that the coefficient of x is −44. 4 More exam questions are available online. 70 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 71 — #17 2.4 The binomial theorem LEARNING INTENTION At the end of this subtopic you should be able to: evaluate expressions using factorial notation calculate combinations expand using the binomial theorem determine particular terms using the binomial theorem. Note: The binomial theorem is not part of the Study Design but is included here to enhance understanding. Pascal’s triangle is useful for expanding small powers of binomial terms. However, to obtain the coefficients required for expansions of higher powers, the triangle needs to be extensively extended. The binomial theorem provides the way around this limitation by providing a rule for the expansion of (x + y)n. Before this theorem can be presented, some notation needs to be introduced. 2.4.1 Factorial notation In this and later topics, calculations such as 7 × 6 × 5 × 4 × 3 × 2 × 1 will be encountered. Such expressions can be written in shorthand as 7! and are read as ‘7 factorial’. There is a factorial key on most calculators. Definition n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1 for any natural number n. It is also necessary to define 0! = 1. 7! is equal to 5040. It can also be expressed in terms of other factorials such as: 7! = 7 × (6 × 5 × 4 × 3 × 2 × 1) 7! = 7 × 6 × (5 × 4 × 3 × 2 × 1) or = 7 × 6! = 7 × 6 × 5! This is useful when working with fractions containing factorials. For example: 7! 7 × 6! 5! 5! = = 6! 6! or 7! 7 × 6 × 5! 1 =7 = 42 By writing the larger factorial in terms of the smaller factorial, the fractions were simplified. Factorial notation is just an abbreviation, so factorials cannot be combined arithmetically. For example, 3! − 2! ≠ 1!. This is verified by evaluating 3! − 2!. 3! − 2! = 3 × 2 × 1 − 2 × 1 = 6−2 =4 ≠1 TOPIC 2 Algebraic foundations 71 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 72 — #18 WORKED EXAMPLE 8 Evaluating an expression with factorial notation 50! Evaluate 5! − 3! +. 49! THINK WRITE 50! 1. Expand the two smaller factorials. 5! − 3! + 49! 50! =5×4×3×2×1−3×2×1+ 49! 50 × 49! 2. To simplify the fraction, write the larger =5×4×3×2×1−3×2×1+ factorial in terms of the smaller factorial. 49! 50 × 49! 3. Calculate the answer. = 120 − 6 + 49! = 120 − 6 + 50 = 164 TI | THINK DISPLAY/WRITE CASIO | THINK DISPLAY/WRITE 1. On a Calculator page, 1. On a Main screen, complete the entry line complete the entry line as: as: 50! 50! 5! − 3! + 5! − 3! + 49! 49! Then press ENTER. Then press EXE. Note: The factorial symbol Note: The factorial symbol is located in CTRL Menu, is located in the Advanced Symbols OR menu of the keyboard. Menu Probability factorial 2. The answer appears on the 164 2. The answer appears on the 164 screen. screen. Resources Resourceseses Interactivity The binomial theorem (int-2555) 2.4.2 Formula for binomial coefficients Each of the terms in the rows of Pascal’s triangle can be expressed using factorial notation. For example, row 3 contains the coefficients 1, 3, 3, 1. 3! 3! 3! 3! These can be written as , , ,. 0! × 3! 1! × 2! 2! × 1! 3! × 0! (Remember that 0! was defined to equal 1.) The coefficients in row 5 (1, 5, 10, 10, 5, 1) can be written as: 5! 5! 5! 5! 5! 5! , , , , , 0! × 5! 1! × 4! 2! × 3! 3! × 2! 4! × 1! 5! × 0! 4! The third term of row 4 would equal and so on. 2! × 2! 72 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 73 — #19 n! The (r + 1)th term of row n would equal. r! × (n − r)! ( ) n n This is normally written using the notations Cr or. r These expressions for the binomial coefficients are referred to as combinatoric coefficients. They occur frequently in other branches of mathematics, including probability theory. Combinatoric coefficients ( ) n n! = = nC r r r! (n − r)! where r ≤ n and r and n are non-negative whole numbers. 2.4.3 Pascal’s triangle with combinatoric coefficients Pascal’s triangle can now be expressed using this notation. ( ) 0 Row 0 0 ( ) ( ) 1 1 Row 1 0 1 ( ) ( ) ( ) 2 2 2 Row 2 0 1 2 ( ) ( ) ( ) ( ) 3 3 3 3 Row 3 0 1 2 3 ( ) ( ) ( ) ( ) ( ) 4 4 4 4 4 Row 4 0 1 2 3 4 Binomial expansions can be expressed using this notation for each of the binomial coefficients. ( ) ( ) ( ) ( ) 3 3 3 3 2 3 2 3 3 The expansion (a + b) = a + a b+ ab + b. 0 1 2 3 Note the following patterns: ( ) ( ) n n =1= (the start and end of each row of Pascal’s triangle) 0 n ( ) ( ) n n =n= (the second from the start and the second from the end of each row) 1 n−1 ( ) ( ) n n =. r n−r TOPIC 2 Algebraic foundations 73 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 74 — #20 WORKED EXAMPLE 9 Evaluating combinations ( ) 8 Evaluate. 3 THINK WRITE ( ) n n! 1. Apply the formula. = r r! (n − r) ! Let n = 8 and r = 3. ( ) 8 8! = 3 3! (8 − 3) ! 8! = 3! 5! 8 × 7 × 6 × 5! 2. Write the largest factorial in terms of the next = largest factorial and simplify. 3! 5! 8×7×6 = 3! 8 × 7 × 6 3. Calculate the answer. = 3 × 2 ×1 =8×7 = 56 2.4.4 The binomial theorem ( ) ( ) ( ) ( ) n n n n The binomial coefficients in row n of Pascal’s triangle can be expressed as , , ,… , and hence 0 1 2 n the expansion of (x + y)n can be formed. The binomial theorem The binomial theorem gives the rule for the expansion of (x + y)n as: ( ) ( ) ( ) n n−1 n n−2 2 n n−r r (x + y) = x + n n x y+ x y +…+ x y + … + yn 1 2 r ( ) ( ) n n since =1=. 0 n n Features of the binomial theorem formula for the expansion of (x + y) There are (n + 1) terms. In each successive term, the powers of x decrease by 1 as the powers of y increase by 1. For each term, the powers of x and y add up to n. For the expansion of (x − y)n , the signs of each term alternate + − + − + … with every even term assigned the − sign and every odd term assigned the + sign. 74 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 75 — #21 WORKED EXAMPLE 10 Expanding using the binomial theorem Use the binomial theorem to expand (3x + 2)4. THINK WRITE 1. Write out the expansion using the (3x + 2)4 ( ) ( ) ( ) binomial theorem. 4 4 3 4 2 2 4 = (3x) + (3x) (2) + (3x) (2) + (3x)(2)3 + (2)4 Note: There should be 5 terms in the 1 2 3 expansion. 2. Evaluate the binomial coefficients. = (3x)4 + 4 × (3x)3 (2) + 6 × (3x)2 (2)2 + 4 × (3x)(2)3 + (2)4 3. Complete the calculations and state = 81x4 + 4 × 27x3 × 2 + 6 × 9x2 × 4 + 4 × 3x × 8 + 16 the answer. ∴ (3x + 2)4 = 81x4 + 216x3 + 216x2 + 96x + 16 2.4.5 Using the binomial theorem The binomial theorem is very useful for expanding (x + y)n. However, for powers n ≥ 7 the calculations can become quite tedious. If a particular term is of interest, forming an expression for the general term of the expansion is an easier option. The general term of the binomial theorem Consider the terms of the expansion: ( ) ( ) ( ) n n n n−1 n n−2 2 n n−r r (x + y) = x + x y+ x y +…+ x y + … + yn 1 2 r ( ) n n 0 Term 1: t1 = xy 0 ( ) n n−1 1 Term 2: t2 = x y 1 ( ) n n−2 2 Term 3: t3 = x y 2 Following the pattern gives: ( ) n n−r r Term (r + 1): tr+1 = x y r The general term of the binomial theorem ( ) n n−r r For the expansion of (x + y) , the general term is tr+1 = n x y. r ( ) n n−r For the expansion of (x − y)n , the general term can be expressed as tr+1 = x (−y)r. r The general term formula enables a particular term to be evaluated without the need to carry out the full expansion. As there are (n + 1) terms in the expansion, if the middle term is sought, there will be two middle terms if n is odd and one middle term if n is even. TOPIC 2 Algebraic foundations 75 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 76 — #22 WORKED EXAMPLE 11 Determining a term in the binomial expansion ( )9 x y Determine the fifth term in the expansion of −. 2 3 THINK WRITE ( )9 x y 1. State the general term formula of the − expansion. 2 3 ( ) ( )n−r ( )r n x y The (r + 1)th term is tr+1 = −. r 2 3 Since the power of the binomial is 9, n = 9. ( ) ( )9−r ( )r 9 x y ∴ tr+1 = − r 2 3 2. Choose the value of r for the required term. For the fifth term, t5 : r+1 = 5 r=4 ( ) ( )9−4 ( )4 9 x y t5 = − 4 2 3 ( ) ( )5 ( )4 9 x y = − 4 2 3 x5 y4 3. Evaluate to obtain the required term. = 126 × × 32 81 7x5 y4 = 144 2.4.6 Identifying a term in the binomial expansion The general term can also be used to determine which term has a specified property, such as the term independent of x or the term containing a particular power of x. WORKED EXAMPLE 12 Determining a particular term 12 Identify which term in the expansion of (8 − 3x2 ) would contain x8 and express the coefficient of x8 as a product of its prime factors. THINK WRITE 12 1. Write down the general term for this (8 − 3x2 ) ) ( expansion. 12 r The general term: tr+1 = (8)12−r (−3x2 ) r ( ) 12 r 2. Rearrange the expression for the general term tr+1 = (8)12−r (−3)r (x2 ) r by grouping the numerical parts together and ( ) the algebraic parts together. 12 = (8)12−r (−3)r x2r r 76 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 77 — #23 3. Find the value of r required to form the given For x8 , 2r = 8, so r = 4. power of x. 4. Identify which term is required. Hence, it is the fifth term that contains x8. 5. Obtain an expression for this term. With r = 4, ( ) 12 t5 = (8)12−4 (−3)4 x8 4 ( ) 12 = (8)8 (−3)4 x8 4 ( ) 12 6. State the required coefficient. The coefficient of x is 8 (8)8 (−3)4. 4 ( ) 12 12 × 11 × 10 × 9 8 7. Express the coefficient in terms of prime (8)8 (−3)4 = × (23 ) × 34 4 4×3×2×1 numbers. = 11 × 5 × 9 × 224 × 34 = 11 × 5 × 32 × 224 × 34 = 11 × 5 × 36 × 224 8. State the answer. Therefore, the coefficient of x8 is 11 × 5 × 36 × 224. 2.4 Exercise Students, these questions are even better in jacPLUS Receive immediate Access Track your feedback and access additional results and sample responses questions progress Find all this and MORE in jacPLUS Technology free 1. Evaluate the following. 7! a. 3! b. 4! c. 5! d. 2! e. 0! × 1! f. 6! 2. Rewrite the following using factorial notation. a. 7 × 6 × 5 × 4 × 3 × 2 × 1 b. 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 c. 8 × 7! d. 9 × 8! 10! 3. WE8 Evaluate 6! + 4! −. 9! n! 4. Simplify. (n − 2)! 5. Evaluate the following. a. 6! b. 4! + 2! 6! c. 7 × 6 × 5! d. 3! TOPIC 2 Algebraic foundations 77 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 78 — #24 6. Evaluate the following. 26! 42! a. b. 24! 43! 49! 69! 11! + 10! c. ÷ d. 50! 70! 11! − 10! ( ) 7 7. WE9 Evaluate. 4 ( ) ( ) n 21 8. Determine an algebraic expression for and use this to evaluate. 2 2 9. Evaluate the following. ( ) ( ) ( ) 5 5 12 a. b. c. 2 3 12 ( ) ( ) 20 7 13 d. C3 e. f. 0 10 10. Simplify the following. ( ) ( ) ( ) n n n+3 a. b. c. 3 n−3 n ( ) ( ) ( ) ( ) 2n + 1 n n n+1 d. e. + f. 2n − 1 2 3 3 11. Simplify the following. n! a. (n + 1) × n! b. (n − 1)(n − 2)(n − 3)! c. (n − 3)! (n − 1)! (n − 1)! (n + 1)! n3 − n2 − 2n (n − 2)! d. e. − f. × (n + 1)! n! (n + 2)! (n + 1)! n−2 Technology active 5 12. WE10 Use the binomial theorem to expand (2x + 3). 7 13. Use the binomial theorem to expand (x − 2). 14. Expand the following. 5 5 6 a. (x + 1) b. (2 − x) c. (2x + 3y) ( )7 ( )8 x 1 10 d. +2 e. x− f. (x2 + 1) 2 x ( )7 x y 15. WE11 Determine the fourth term in the expansion of −. 3 2 ( )10 2y 16. Determine the middle term in the expansion of x +. 2 8 17. WE12 Identify which term in the expansion of (4 + 3x3 ) would contain x15 and express the coefficient of 15 x as a product of its prime factors. ( )6 2 18. Determine the term independent of x in the expansion of x+. x 78 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 79 — #25 For questions 19 and 20, use either expansion or the general term formula. 19. Obtain each of the following terms. 6 a. The fourth term in the expansion of (5x + 2) 6 b. The third term in the expansion of (3x2 − 1) 7 c. The middle term(s) in the expansion of (x + 2y) 9 20. a. Obtain the coefficient of x6 in the expansion of (1 − 2x2 ). 11 b. Express the coefficient of x5 in the expansion of (3 + 4x) as a product of its prime factors. ( )10 1 c. Determine the term independent of x in the expansion of x2 +. x3 21. Evaluate the following using CAS technology. ( ) 15 a. 15! b. 10 ( ) n 22. a. Solve for n: = 1770. 2 ( ) 12 b. Solve for r: = 220. r 2.4 Exam questions Question 1 (1 mark) TECH-ACTIVE 4! + 3! − 2! MC The value of is 4! − 3! + 2! A. 1 B. −1 C. 20 7 D. 5 7 E. − 5 Question 2 (1 mark) TECH-ACTIVE ( ) ( ) 5 4 MC The value of 3 × +4× is 3 2 A. 13 B. 16 C. 21 D. 54 E. 78 Question 3 (4 marks) TECH-ACTIVE ( )7 Identify which term in the expansion of 3 − 4x3 would contain x12 and express the coefficient of x12 as a product of its prime factors. More exam questions are available online. TOPIC 2 Algebraic foundations 79 “c02AlgebraicFoundations_PrintPDF” — 2022/11/22 — 14:20 — page 80 — #26 2.5 Sets of real numbers LEARNING INTENTION At the end of this subtopic you should be able to: classify numbers as elements of the real number system use interval notation and number lines to describe sets. The concept of numbers in counting and the introduction of symbols for numbers marked the beginning of a major intellectual development in the minds of humans. Every civilisation appears to have developed a system for counting using written or spoken symbols for numbers. Over time, technologies were devised to assist in counting and computational techniques, and from these counting machines the computer was developed. Over the course of history, different categories of numbers have been defined that collectively form the real number system. Real numbers are all the numbers that are positive, zero or negative. Before further describing and classifying the real number system, a review of some mathematical notation is given. Resources Resourceseses Interactivity Sets (int-2556) 2.5.1 Set notation A set is a collection of objects, these objects being referred to as the elements of the set. For example, if the universal set is 𝜖 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then sets could include set A = {1, 2, 3, 4, 5}, set B = {1, 3, 5} and set C = {2, 4, 6, 8, 10}. The table below gives the common symbols used in set notation. Symbol Example Meaning ∈ 2∈A 2 is an element of set A, or belongs to set A. ∉ 2∉B 2 is not an element of set B, or does not belong to set B. ⊂ B⊂A Set B is a subset of set A, or every element of set B is an element of set A. ⊄ A⊄B Set A is not a subset of set B. ∪ A∪C The union of sets A and C; contains the elements that are either in A or in C or in both, giving A ∪ C = {1, 2, 3, 4, 5, 6, 8, 10} ∩ A∩C The intersection of sets A and C; contains the elements that are in both A and C, giving A ∩ C = {2, 4} \ A\C The exclusion symbol; the elements of set A that are not in set C, giving A\C = {1, 3, 5} ∅ B∩C=∅ The empty set, or disjoint sets, since sets B and C do not have any common elements. ′ A ∩ A′ = ∅ The complement of A; the set of elements not in A, with the symbol A′ , giving A′ = {6, 7, 8, 9, 10} 80 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 81 — #27 2.5.2 Classification of numbers Although counting numbers are sufficient to solve equations such as 2 + x = 3, they are not sufficient to solve, for example, 3 + x = 2, where negative numbers are needed, or 3x = 2, where fractions are needed. The table below shows the classification of numbers in our real number system. N = {1, 2, 3, 4, …} The set of natural numbers, that is the positive whole numbers or counting numbers Z = {… − 2, −1, 0, 1, 2 …} The set of integers, that is all positive and negative whole numbers and the number zero −9 The set of rational numbers, all numbers that can be expressed in the Q = {… , 0.75, 0.3, 5,..} p 8 form , q ≠ 0 and p, q ∈ Z, that is all fractions, integers and finite and q recurring decimals. √ √ I = {… − 3,.., 𝜋,.. 2,..} The set of irrational numbers where Q ∩ I = ∅ or I = R\Q, that is all real numbers that are not rational. Some irrational numbers, such as 𝜋 and e, are called transcendental numbers. R=Q∪ I The set of real numbers, that is the union of the set of rational numbers and the set of irrational numbers The relationship between the sets, N ⊂ Z ⊂ Q ⊂ R, can be illustrated as shown. R N = natural numbers Q Z = integers Z I Q = rational numbers N I = irrational numbers R = real numbers The set of all real numbers forms a number line continuum on which all of the positive, zero or negative numbers are placed. Hence, R = R− ∪ {0} ∪ R+. R– Zero R+ 0 The sets of the real number system can also be viewed as the following hierarchy. Real numbers R Irrational numbers I Rational numbers Q (surds, non-terminating and non-recurring decimals, π, e) Non-integer rationals Integers (terminating and Z recurring decimals) TOPIC 2 Algebraic foundations 81 “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 82 — #28 Expressions and symbols that do not represent real numbers It is important to recognise that the following are not numbers. The symbol for infinity ∞ may suggest this is a number, but that is not so. We can speak of numbers getting larger and larger and approaching infinity, but infinity is a concept, not an actual number. a Any expression of the form does not represent a number, since division by zero is not possible. If a = 0, 0 0 the expression is said to be indeterminate. It is not defined as a number. 0 This is beyond the Mathematical Methods course, but there are numbers √ that are not elements of the set of real numbers. For example, the square roots of negative numbers, such as −1, are unreal, but these square roots are numbers. They belong to the set of complex numbers. These numbers are very important in higher levels of mathematics. WORKED EXAMPLE 13 Classifying numbers as elements of the real number system a. Classify each of the following numbers as an element of a subset of the real numbers. 3 √ √ i. − ii. 7 iii. 6 − 2 × 3 iv. 9 5 b. Identify which of the following are correct statements. i. 5 ∈ Z ii. Z ⊂ N iii. R− ∪ R+ = R THINK WRITE 3 a. i. Fractions are rational numbers. a. i. − ∈Q 5 √ ii. Surds are irrational numbers. ii. 7∈I iii. Evaluate the number using the correct iii. 6 − 2 × 3 = 6 − 6 order of operations. =0 ∴ (6 − 2 × 3) ∈ Z √ iv. Evaluate the square root. iv. 9=3 √ ∴ 9∈Z b. i. Z is the set of integers. b. i. 5 ∈ Z is a correct statement since 5 is an integer. ii. N is the set of natural or counting ii. Z ⊂ N is incorrect since N ⊂ Z. numbers. iii. This is the union of R− , the set of negative iii. R− ∪ R+ = R is incorrect since R includes real numbers, and R+ , the set of positive the number zero, which is neither positive real numbers. nor negative. 2.5.3 Interval notation Interval notation provides an alternative and often convenient way of describing certain sets of numbers. Closed interval [a, b] = {x ∶ a ≤ x ≤ b} is the set of real numbers that lie between a and b, including the end points, a and b. The inclusion of the end points is indicated by the use of the square brackets []. 82 Jacaranda Maths Quest 11 Mathematical Methods VCE Units 1 & 2 Third Edition “c02AlgebraicFoundations_PrintPDF” — 2022/8/1 — 14:30 — page 83 — #29 This is illustrated on a number line using closed circles at the end points. a b Open interval (a, b) = {x ∶ a < x < b} is the set of real numbers that lie between a and b, not including the end points, a and b. The exclusion of the end points is indicated by the use of the round brackets (). This is illustrated on a number line using open circles at the end points. a b Half-open intervals Half-open intervals have only one end point included. [a, b) = {x ∶ a ≤ x < b} a b (a, b] = {x ∶ a < x ≤ b} a b Interval notation can be used for infinite intervals using the symbol for infinity with an open end. For example, the set of real numbers, R, is the same as the interval (−∞, ∞). WORKED EXAMPLE 14 Illustrating a set on the number line a. Illustrate each of the following sets on a number line and express each set in alternative notation. i. (−2, 2] ii. {x ∶ x

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