2Algebraic. foundations

Transcription

1 2Algebraic foundations 2. Kick off with CAS 2.2 Algebraic skills 2.3 Pascal s triangle and binomial expansions 2.4 The Binomial theorem 2.5 Sets of real numbers 2.6 Surds 2.7 Review c02algebraicfoundations.indd 58 7//5 3: AM

2 2. Kick off with CAS playing lotto Using technology calculate the following products. a 3 2 b c d Using CAS technology, find the symbol! and evaluate the following: a 3! b 5! c 7! d 0! This symbol is called factorial. Compare the answers to questions and 2. Using the factorial symbol or another method, how many ways can a one number be arranged? b two numbers be arranged? c three numbers be arranged? d six numbers be arranged e nine numbers be arranged? Using the factorial symbol or another method, a how many 2 digit numbers can be formed from 6 different numbers? b how many 3 digit numbers can be formed from 5 different numbers? c how many 4 digit numbers can be formed from 0 different numbers? In the game of lotto, how many different combinations of 6 numbers can be chosen from 45 numbers? Please refer to the Resources tab in the Prelims section of your ebookplus for a comprehensive step-by-step guide on how to use your CAS technology. c02algebraicfoundations.indd 59 7//5 3: AM

3 Unit & 2 AOS 2 Topic Concept 2.2 Algebraic skills Concept summary Algebraic skills This chapter covers some of the algebraic skills required as the foundation to learning and understanding the functionality concepts we will encounter in the study of Mathematical Methods. Some basic algebraic techniques will be revised and some new techniques will be introduced. review of factorisation and expansion expansion The Distributive Law a(b + c) ab + ac is fundamental in expanding to remove brackets. Some simple expansions include: (a + b)(c + d) ac + ad + bc + bd (a + b) 2 (a + b)(a + b) a 2 + 2ab + b 2 (a b) 2 a 2 2ab + b 2 (a + b)(a b) a 2 b 2 Worked example Expand 2(4x 3) 2 (x 2)(x + 2) + (x + 5)(2x ) and state the coefficient of the term in x. THINK Expand each pair of brackets. Note: The first term contains a perfect square, the second a difference of two squares and the third a quadratic trinomial. WRITE 2(4x 3) 2 (x 2)(x + 2) + (x + 5)(2x ) 2(6x 2 24x + 9) (x 2 4) + (2x 2 x + 0x 5) 2 Expand fully, taking care with signs. 32x 2 48x + 8 x x 2 + 9x 5 3 Collect like terms together. 33x 2 39x State the answer. Note: Read the question again to ensure the answer given is as requested. The expansion gives 33x 2 39x + 7 and the coefficient of x is 39. factorisation Some simple factors include: common factor difference of two perfect squares perfect squares and factors of other quadratic trinomials. expand Factorised form (a + b)(c + d) is equal to factorise ac + ad + bc + bd Expanded form 60 maths QuesT mathematical methods Vce units and 2 c02algebraicfoundations.indd 60 7//5 3: AM

4 A systematic approach to factorising is displayed in the following diagram. Common factor? ac + ad a(c + d) How many terms? Two terms Three terms Four or more terms Difference of two squares? a 2 b 2 (a + b)(a b) Quadratic trinomial? a 2 + 3ab + 2b 2 (a + 2b)(a + b) a 2 + 2ab + b 2 (a + b) 2 a 2 2ab + b 2 (a b) 2 Grouping? ac + ad + bc + bd a(c + d) + b(c + d) (a + b)(c + d) Grouping terms commonly referred to as grouping 2 and 2 and grouping 3 and depending on the number of terms grouped together, are often used to create factors. For example, as the first three terms of a 2 + 2ab + b 2 c 2 are a perfect square, grouping 3 and would create a difference of two squares expression, allowing the whole expression to be factorised. a 2 + 2ab + b 2 c 2 (a 2 + 2ab + b 2 ) c 2 (a + b) 2 c 2 This factorises to give (a + b c)(a + b + c). Worked example 2 Factorise: a 2x 3 + 5x 2 y 2y 2 x b 4y 2 x 2 + 0x 25 c 7(x + ) 2 8(x + ) + using the substitution a (x + ). THINK WRITE a Take out the common factor. a 2x 3 + 5x 2 y 2y 2 x x(2x 2 + 5xy 2y 2 ) 2 Factorise the quadratic trinomial. x(2x 3y)(x + 4y) b The last three terms of the expression can be grouped together to form a perfect square. b 4y 2 x 2 + 0x 25 4y 2 (x 2 0x + 25) Topic 2 Algebraic foundations 6 c02algebraicfoundations.indd 6 7//5 3: AM

5 2 Use the grouping 3 and technique to create a difference of two squares. 4y 2 (x 5) 2 (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 answer. (2y x + 5)(2y + x 5) c Substitute a (x + ) to form a quadratic trinomial in a. c 7(x + ) 2 8(x + ) + 7a 2 8a + where a (x + ) 2 Factorise the quadratic trinomial. (7a )(a ) 3 Substitute (x + ) back in place of a (7(x + ) ) ( (x + ) ) and simplify. 4 Remove the inner brackets and simplify to obtain the answer. (7x + 7 )(x + ) (7x + 6)(x) x(7x + 6) Factorising sums and differences of two perfect cubes Check the following by hand or by using a CAS technology. Expanding (a + b)(a 2 ab + b 2 ) gives a 3 + b 3, the sum of two cubes; and expanding (a b)(a 2 + ab + b 2 ) gives a 3 b 3, the difference of two cubes. Hence the factors of the sum and difference of two cubes are: a 3 + b 3 (a + b)(a 2 ab + b 2 ) a 3 b 3 (a b)(a 2 + ab + b 2 ) Worked example 3 Factorise: a x 3 27 b 2x THINK WRITE a Express x 3 27 as a difference of two cubes. a x 3 27 x Apply the factorisation rule for the difference of two cubes. Using a 3 b 3 (a b)(a 2 + ab + b 2 ) with a x, b 3, x (x 3)(x 2 + 3x ) 3 State the answer. x 3 27 (x 3)(x 2 + 3x + 9) b Take out the common factor. b 2x (x 3 + 8) 2 Express x as a sum of two cubes. 2(x ) 3 Apply the factorisation rule for the sum of two cubes. Using a 3 + b 3 (a + b)(a 2 ab + b 2 ) with a x, b 2, x (x + 2)(x 2 2x ) 2(x ) 2(x + 2)(x 2 2x ) 4 State the answer. 2x (x + 2)(x 2 2x + 4) 62 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 62 7//5 3: AM

6 Algebraic fractions Factorisation techniques may be used in the simplification of algebraic fractions under the arithmetic operations of multiplication, division, addition and subtraction. Multiplication and division of algebraic fractions An algebraic fraction can be simplified by cancelling any common factor between its numerator and its denominator. For example: ab + ac ad a(b + c) ad b + c d 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: a(b + c) ad d(a + c) b (b + c)(a + c) b 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 b c d a b d c Worked example 4 Simplify a x 2 2x x 2 5x + 6 b x4 x 3 + x2 3 x THINK a Factorise both the numerator and the denominator. Note: The numerator has a common factor; the denominator is a quadratic trinomial. WRITE a x 2 2x x 2 5x + 6 x(x 2) (x 3)(x 2) 2 Cancel the common factor in the numerator and denominator. x(x 2) (x 3)(x 2) 3 Write the fraction in its simplest form. x x 3 No further cancellation is possible. b Change the division into multiplication by replacing the divisor by its reciprocal. b x4 x 3 + x2 3 x x4 x 3 3 x + x 2 Topic 2 Algebraic foundations 63 c02algebraicfoundations.indd 63 7//5 3: AM

7 2 Factorise where possible. Note: The aim is to create common factors of both the numerator and denominator. For this reason, write (3 x) as (x 3). 3 Cancel the two sets of common factors of the numerator and denominator. 4 Multiply the remaining terms in the numerator together and the remaining terms in the denominator together. 5 State the answer. Note: The answer could be expressed in different forms, including as a product of linear factors, but this is not necessary as it does not lead to any further simplification. Since, x 4 (x 2 ) 2 2 (x 2 )(x 2 + ) Then: x 4 x 3 3 x + x 2 (x2 )(x 2 + ) x 3 (x2 )(x 2 + ) x 3 (x2 ) (x2 ) (x 2 ) x 2 (x 3) + x 2 (x 3) + x 2 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 Simplify 2 3x + 3 x 2 + x x 2 x 2. THINK Factorise each denominator. 2 Select the lowest common denominator and express each fraction with this as its denominator. WRITE 2 3x + 3 x 2 + x x 2 x 2 2 3(x + ) (x 2) + x (x + )(x 2) 2 (x 2) 3(x + )(x 2) 3(x + ) 3(x + )(x 2) + x 3 3(x + )(x 2) 64 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 64 7//5 3:2 AM

8 2(x 2) 3(x + ) + 3x 3 Combine the fractions into one fraction. 3(x + )(x 2) 4 Expand the terms in the numerator. Note: It is not necessary to expand the denominator terms. 5 Collect like terms in the numerator and state the answer. Note: Since there are no common factors between the numerator and the denominator, the fraction is in its simplest form. 2x 4 3x 3 + 3x 3(x + )(x 2) 2x 7 3(x + )(x 2) Exercise 2.2 PRactise Consolidate Algebraic skills WE Expand 3(2x + ) 2 + (7x + )(7x ) (3x + 4)(2x ) and state the coefficient of the term in x. 2 Expand (2 + 3x)(x + 6)(3x 2)(6 x). 3 WE2 Factorise: a 4x 3 8x 2 y 2y 2 x b 9y 2 x 2 8x 6 c 4(x 3) 2 3(x 3) 22 using the substitution a (x 3). 4 Factorise x 2 6x + 9 xy + 3y. 5 WE3 Factorise the following. a x 3 25 b 3 + 3x 3 6 Factorise 2y 4 + 2y(x y) 3. 7 WE4 Simplify: x a 2 + 4x x 2 + 2x 8 8 Simplify x3 25 x x 3 + 5x x. 9 WE5 Simplify 6 5x 25 + x 2x x 2 6x Simplify a 4 x + 3 (x + ) 2b 6x2 x 2 + 2x +. b x x x2 + 8 x 5. This exercise should be attempted by hand rather than by using a CAS technology. Expand each of the following expressions. a (2x + 3) 2 b 4a(b 3a)(b + 3a) c 0 (c + 2)(4c 5) d (5 7y) 2 e (3m 3 + 4n)(3m 3 4n) f (x + ) 3 2 Expand the following. a (g h) 2 b (2p + 7q) 2 (7q 2p) c (x + 0)(5 + 2x)(0 x)(2x 5) Topic 2 Algebraic foundations 65 c02algebraicfoundations.indd 65 7//5 3:2 AM

9 3 Expand and simplify, and state the coefficient of the term in x. a 2(2x 3)(x 2) + (x + 5)(2x ) b (2 + 3x)(4 6x 5x 2 ) (x 6)(x + 6) c (4x + 7)(4x 7)( x) d (x + 2y)(x + + 2y) + (x ) 2 e (3 2x)(2x + 9) 3(5x )(4 x) f x 2 + x 4(x 2 + x 4) 4 Factorise each of the following expressions. a x 2 + 7x 60 b 4a 2 64 c 2bc + 2b + + c d 5x x 2 e 9( m) 2 f 8x 2 48xy + 72y 2 5 Fully factorise the following. a x 3 + 2x 2 25x 50 b 00p 3 8pq 2 c 4n 2 + 4n + 4p 2 d 49(m + 2n) 2 8(2m n) 2 e 3(a ) + 52( a) 3 f a 2 b 2 a + b + (a + b ) 2 6 Use a substitution method to factorise the following. a (x + 5) 2 + (x + 5) 56 b 2(x + 3) 2 7(x + 3) 9 c 70(x + y) 2 y(x + y) 6y 2 d x 4 8x 2 9 e 9(p q) 2 + 2(p 2 q 2 ) + 4(p + q) 2 f a 2 aa + a b 2 4a 2 aa + a b + 4a2 7 Factorise the following. a x 3 8 b x c x 3 d 27x y 3 e x 4 25x f (x ) Fully factorise the following. a 24x 3 8y 3 b 8x 4 y 4 + xy c 25(x + 2) (x 5) 3 d 2(x y) 3 54(2x + y) 3 e a 5 a 3 b 2 + a 2 b 3 b 5 f x 6 y 6 9 Simplify the following. a 3x2 7x x 2 b x3 + 4x 2 9x 36 x 2 + x 2 c (x + h)3 x 3 h e m3 2m 2 n m 2 4n 2 m 3 + n 3 m 2 + 3mn + 2n 2 d 2x 2 9x 3 + 3x 9x 2 2 8x 2 2x + 2 f x3 + x x2 3 + x + x + x2 2 x + x 2 66 maths QuesT mathematical methods Vce units and 2 c02algebraicfoundations.indd 66 7//5 3:2 AM

10 MasTER 20 Simplify the following expressions. 4 a x x x 2 b c d e 4 x x x 2 5 x x + 3 x 2 + x 30 4y 2 36y y 2 8 2y 2 9y p q p p 2 q q 3 2 p 4 q 4 7 f (a + 6b) a a 2 3ab + 2b 5 2 a 2 ab 2b 2b 2 a Expand (x + 5)(2 x)(3x + 7). b Factorise 27(x 2) (x + 2) a Simplify 3 x + 8 x + 8. b Use CAS technology to complete some of the questions in Exercise 2.2. The word algebra is of arabic origin. It is derived from al jabr, and was developed by the mathematician Muhammad ibn Musa al Khwarizmi (c ). The word algorithm is derived from his name. 2.3 Unit & 2 AOS 2 Topic Concept 2 Pascal s triangle and binomial expansions Concept summary Pascal s triangle and binomial expansions expansions of perfect cubes The perfect square (a + b) 2 may be expanded quickly by the rule (a + b) 2 a 2 + 2ab + b 2. 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)(a 2 + 2ab + b 2 ) a 3 + 2a 2 b + ab 2 + ba 2 + 2ab 2 + b 3 a 3 + 3a 2 b + 3ab 2 + b 3 Therefore, the rules for expanding a perfect cube are: (a + b) 3 a 3 + 3a 2 b + 3ab 2 + b 3 (a b) 3 a 3 3a 2 b + 3ab 2 b 3. 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, 3, 3,. The coefficients of each term in the expansion of (a b) 3 are, 3, 3,. The signs alternate + + in the expansion of (a b) 3. Topic 2 AlgebrAic foundations 67 c02algebraicfoundations.indd 67 7//5 3:2 AM

11 Worked example 6 Expand (2x 5) 3. THINK WRITE Use the rule for expanding a perfect cube. (2x 5) 3 Using (a b) 3 a 3 3a 2 b + 3ab 2 b 3, 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. 8x 3 3 4x x x 3 60x x 25 3 State the answer. (2x 5) 3 8x 3 60x x 25 4 An alternative approach to using the rule would be to write the expression in the form (a + b) 3. (2x 5) 3 (2x + ( 5)) 3 Using (a + b) 3 a 3 + 3a 2 b + 3ab 2 + b 3, let 2x a and 5 b. (2x 5) 3 (2x + ( 5) ) 3 (2x) 3 + 3(2x) 2 ( 5) + 3(2x)( 5) 2 + ( 5) 3 8x 3 60x x 25 Interactivity Pascal s triangle int-2554 Pascal s triangle Although known to Chinese mathematicians many centuries earlier, the following pattern is named after the seventeenth century French mathematician Blaise Pascal. Pascal s triangle contains many fascinating patterns. Each row from row onwards begins and ends with. Each other number along a row is formed by adding the two terms to its left and right from the preceding row. Row 0 Row Row 2 2 Row Row The numbers in each row are called binomial coefficients. 68 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 68 7//5 3:2 AM

12 The numbers, 2, in row 2 are the coefficients of the terms in the expansion of (a + b) 2. (a + b) 2 a 2 + 2ab + b 2 The numbers, 3, 3, in row 3 are the coefficients of the terms in the expansion of (a + b) 3. (a + b) 3 a 3 + 3a 2 b + 3ab 2 + b 3 Each row of Pascal s triangle contains the coefficients in the expansion of a power of (a + b). Such expansions are called binomial expansions because of the two terms a and b in the brackets. Row n contains the coefficients in the binomial expansion (a + b) n. To expand (a + b) 4 we would use the binomial coefficients,, 4, 6, 4,, from row 4 to obtain: (a + b) 4 a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 Notice that the powers of a decrease by as the powers of b increase by, 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 a 4 4a 3 b + 6a 2 b 2 4ab 3 + b 4 By extending Pascal s triangle, higher powers of such binomial expressions can be expanded. Worked example 7 Form the rule for the expansion of (a b) 5 and hence expand (2x ) 5. THINK Choose the row in Pascal s triangle which contains the required binomial coefficients. 2 Write down the required binomial expansion. 3 State the values to substitute in place of a and b. 4 Write down the expansion. 5 Evaluate the coefficients and state the answer. WRITE For (a b) 5, the power of the binomial is 5. Therefore the binomial coefficients are in row 5. The binomial coefficients are:, 5, 0, 0, 5, Alternate the signs: (a b) 5 a 5 5a 4 b + 0a 3 b 5 0a 2 b 3 + 5ab 4 b 5 To expand (2x ) 5, replace a by 2x and b by. (2x ) 5 (2x) 5 5(2x) 4 () + 0(2x) 3 () 2 0(2x) 2 () 3 + 5(2x)() 4 () 5 32x 5 5 6x x 3 0 4x 2 + 0x 32x 5 80x x 3 40x 2 + 0x (2x ) 5 32x 5 80x x 3 40x 2 + 0x Topic 2 Algebraic foundations 69 c02algebraicfoundations.indd 69 7//5 3:2 AM

13 Exercise 2.3 PRactise Consolidate Pascal s triangle and binomial expansions WE6 Expand (3x 2) 3. 2 Expand a a b2 b and give the coefficient of a 2 b 2. 3 WE7 Form the rule for the expansion of (a b) 6 and hence expand (2x ) 6. 4 Expand (3x + 2y) 4. 5 Expand the following. a (3x + ) 3 b ( 2x) 3 c (5x + 2y) 3 d a x 2 y 3 b 3 6 Select the correct statement(s). a (x + 2) 3 x 3 + 6x 2 + 2x + 8 b (x + 2) 3 x c (x + 2) 3 (x + 2)(x 2 2x + 4) d (x + 2) 3 (x + 2)(x 2 + 2x + 4) e (x + 2) 3 x 3 + 3x 2 + 3x Find the coefficient of x 2 in the following expressions. a (x + ) 3 3x(x + 2) 2 b 3x 2 (x + 5)(x 5) + 4(5x 3) 3 c (x )(x + 2)(x 3) (x ) 3 d (2x 2 3) 3 + 2(4 x 2 ) 3 8 a Write down the numbers in row 7 of Pascal s triangle. b Identify which row of Pascal s triangle contains the binomial coefficients:, 9, 36, 84, 26, 26, 84, 36, 9,. c Row 0 contains term, row contains 2 terms. How is the number of terms related to the row number of Pascal s triangle? 9 Copy and complete the following table by making use of Pascal s triangle. Binomial power Expansion Number of terms in the expansion Sum of indices in each term (x + a) 2 (x + a) 3 (x + a) 4 (x + a) 5 0 Expand the following using the binomial coefficients from Pascal s triangle. a (x + 4) 5 b (x 4) 5 c (xy + 2) 5 d (3x 5y) 4 e (3 x 2 ) 4 f ( + x) 6 ( x) 6 a Expand and simplify 3 (x ) + y4 4. b Find the term independent of x in the expansion of a x x b 6. c If the coefficient of x 2 y 2 in the expansion of (x + ay) 4 is 3 times the coefficient of x 2 y 3 in the expansion of (ax 2 y) 4, find the value of a. d Find the coefficient of x in the expansion of ( + 2x)( x) Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 70 7//5 3:2 AM

14 2 a The sum of the binomial coefficients in row 2 is Form the sum of the binomial coefficients in each of rows 3 to 5. b Create a formula for the sum of the binomial coefficients of row n. c Expand ( + x) 4. d In the expansion in part c, let x and comment on the result. e Using a suitably chosen value for x evaluate. 4 using the expansion in part c. 3 a Expand (x + ) 5 (x + ) 4 and hence show that (x + ) 5 (x + ) 4 x(x + ) 4. b Prove (x + ) n+ (x + ) n x(x + ) n. 4 A section of Pascal s triangle is shown. Determine the values of a, b and c. 45 a b c Master 5 Pascal s triangle can be written as: a Describe the pattern in the second column. b What would be the sixth entry in the third column? c Describe the pattern of the terms in the third column by forming a rule for the nth entry. d What would be the rule for finding the nth entry of the fourth column? 6 Expand ( + x + x 2 ) 4 and hence, using a suitably chosen value for x, evaluate The Pascal triangle as depicted in 303 in a work by the Chinese mathematician Chu Shih Chieh. Topic 2 Algebraic foundations 7 c02algebraicfoundations.indd 7 7//5 3:2 AM

15 2.4 Unit & 2 AOS 2 Topic Concept 3 Binomial theorem Concept summary Interactivity Binomial theorem int-2555 The Binomial theorem 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. Factorial notation In this and later chapters, calculations such as 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, but it is advisable to remember some small factorials by heart. Definition n! n (n ) (n 2) for any natural number n. It is also necessary to define 0!. 7! is equal to It can also be expressed in terms of other factorials such as: 7! ! or 7 6! 7 6 5! and other variations. This is useful when working with fractions containing factorials. For example: 7! 6! 7 6! 5! 6! 7! 5! 7 6 5! or 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!!. This is verified by evaluating 3! 2!. 3! 2! Worked example 8 Evaluate 5! 3! + 50! 49! THINK WRITE Expand the two smaller factorials. 5! 3! + 50! 49! ! 49! 72 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 72 7//5 3:2 AM

16 2 To simplify the fraction, write the larger factorial in terms of the smaller factorial ! 49! Calculate the answer 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, 3, 3,. 3! These can be written as 0! 3!, 3!! 2!, 3! 2!!, 3! 3! 0!. (Remember that 0! was defined to equal ). The coefficients in row 5 (, 5, 0, 0, 5, ) can be written as: 5! 0! 5!, 5!! 4!, 5! 2! 3!, 5! 3! 2!, 5! 4!!, 5! 5! 0! 4! The third term of row 4 would equal and so on. 2! 2! n! The (r + ) th term of row n would equal. This is normally written r! (n r)! using the notations n C r or a n r b. These expressions for the binomial coefficients are referred to as combinatoric coefficients. They occur frequently in other branches of mathematics including probability theory. Blaise Pascal is regarded as the father of probability and it could be argued he is best remembered for his work in this field. It is not necessary to include the multiplication sign between factorials. Pascal s triangle with combinatoric coefficients Pascal s triangle can now be expressed using this notation: 50 49! 49! a n r b n! r!(n r)! n C r, where r n and r, n are non-negative whole numbers. Row 0 a 0 0 b Row a 0 b a b Row 2 a 2 0 b a 2 b a2 2 b Row 3 a 3 0 b a3 b a3 2 b a3 3 b Row 4 a 4 0 b a4 b a4 2 b a4 3 b a4 4 b Topic 2 Algebraic foundations 73 c02algebraicfoundations.indd 73 7//5 3:2 AM

17 Binomial expansions can be expressed using this notation for each of the binomial coefficients. The expansion (a + b) 3 a 3 0 ba3 + a 3 ba2 b + a 3 2 bab2 + a 3 3 bb3. Note the following patterns: a n 0 b an b (the start and end of each row of Pascal s triangle) n a n b n a n b (the second from the start and the second from the end of each n row) and a n r b a n n r b. While most calculators have a n C r key to assist with the evaluation of the coefficients, the formula for a n r b or n C r should be known. Some values of a n b can be committed to memory. r Worked example 9 Evaluate a 8 3 b. THINK WRITE Apply the formula. a n r b n! r!(n r)! Let n 8 and r 3. a 8 3 b 8! 3!(8 3)! 8! 3!5! 2 Write the largest factorial in terms of the next largest ! factorial and simplify. 3!5! ! 3 Calculate the answer Binomial theorem The binomial coefficients in row n of Pascal s triangle can be expressed as a n 0 b, an b, an 2 b,...an n b and hence the expansion of (x + y)n can be formed. The binomial theorem gives the rule for the expansion of (x + y) n as: (x + y) n a n 0 bxn + a n bxn y + a n 2 bxn 2 y a n r bxn r y r a n n byn 74 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 74 7//5 3:2 AM

18 Since a n 0 b an b this formula becomes: n (x + y) n x n + a n bxn y + a n 2 bxn 2 y a n r bxn r y r + + y n Features of the binomial theorem formula for the expansion of (x + y) n There are (n + ) terms. In each successive term the powers of x decrease by as the powers of y increase by. For each term, the powers of x and y add up to n. For the expansion of (x y) n the signs alternate with every even term assigned the sign and every odd term assigned the + sign. Worked example 0 Use the binomial theorem to expand (3x + 2) 4. THINK Write out the expansion using the binomial theorem. Note: There should be 5 terms in the expansion. 2 Evaluate the binomial coefficients. 3 Complete the calculations and state the answer. WRITE (3x + 2) 4 (3x) 4 + a 4 b (3x) 3 (2) + a 4 2 b (3x) 2 (2) 2 + a 4 3 b (3x) (2) 3 + (2) 4 (3x) (3x) 3 (2) + 6 (3x) 2 (2) (3x)(2) 3 + (2) 4 8x x x x (3x + 2) 4 8x x x x + 6 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 then, rather than expand the expression completely to obtain the desired term, an alternative option is to form an expression for the general term of the expansion. The general term of the binomial theorem Consider the terms of the expansion: (x + y) n x n + a n bxn y + a n 2 bxn 2 y a n r bxn r y r y n Term : t a n 0 bxn y 0 Term 2: t 2 a n bxn y Term 3: t 3 a n 2 bxn 2 y 2 Topic 2 Algebraic foundations 75 c02algebraicfoundations.indd 75 7//5 3:2 AM

19 Following the pattern gives: Term (r + ): t r+ a n r bxn r y r For the expansion of (x + y) n, the general term is t r+ a n r bxn r y r. For the expansion of (x y) n, the general term could be expressed as t r+ a n r bxn r ( y) 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 + ) 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. Worked example Find the fifth term in the expansion of a x 2 y 3 b 9. THINK WRITE State the general term formula of the expansion. a x 2 y 9 3 b The (r + )th term is t r+ a n r b ax 2 b n r a y 3 b r. Since the power of the binomial is 9, n 9. t r+ a 9 r b ax 2 b 9 r a y 3 b r 2 Choose the value of r for the required term. For the fifth term, t 5 : r + 5 r 4 t 5 a 9 4 b ax 2 b 9 4 a y 3 b 4 a 9 4 b a x 2 b 5 a y 3 b 4 3 Evaluate to obtain the required term. 26 x5 32 y4 8 7x5 y 4 44 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. 76 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 76 7//5 3:2 AM

20 Worked example 2 Identify which term in the expansion of (8 3x 2 ) 2 would contain x 8 and express the coefficient of x 8 as a product of its prime factors. THINK WRITE Write down the general term for this expansion. (8 3x 2 ) 2 2 Rearrange the expression for the general term by grouping the numerical parts together and the algebraic parts together. 3 Find the value of r required to form the given power of x. The general term: t r+ a 2 t r+ a 2 r b (8) 2 r ( 3) r (x 2 ) r a 2 r b (8) 2 r ( 3) r x 2r For x 8, 2r 8 so r 4. r b(8)2 r ( 3x 2 ) r 4 Identify which term is required. Hence it is the fifth term which contains x 8. 5 Obtain an expression for this term. With r 4, t 5 a 2 4 b (8) 2 4 ( 3) 4 x 8 a 2 4 b (8) 8 ( 3) 4 x 8 6 State the required coefficient. The coefficient of x 8 is a 2 4 b(8)8 ( 3) 4. 7 Express the coefficient in terms of prime numbers. a 2 4 b(8)8 ( 3) (2 3 ) State the answer. Therefore the coefficient of x 8 is Exercise 2.4 PRactise The Binomial theorem WE8 Evaluate 6! + 4! 0! 9!. 2 Simplify n! (n 2)!. 3 WE9 Evaluate a 7 4 b. 4 Find an algebraic expression for a n b and use this to evaluate a2 2 2 b. Topic 2 Algebraic foundations 77 c02algebraicfoundations.indd 77 7//5 3:2 AM

21 5 WE0 Use the binomial theorem to expand (2x + 3) 5. 6 Use the binomial theorem to expand (x 2) 7. 7 WE Find the fourth term in the expansion of a x 3 y 2 b 7. 8 Find the middle term in the expansion of ax 2 + y 2 b 0. 9 WE2 Identify which term in the expansion of (4 + 3x 3 ) 8 would contain x 5 and express the coefficient of x 5 as a product of its prime factors. 0 Find the term independent of x in the expansion of ax + 2 x b 6. Consolidate Evaluate the following. a 6! b 4! + 2! c 7 6 5! d 6! 3! e 0! 9! f (4! + 3!) 2 2 Evaluate the following. a 26! b 42! c 49! 24! 43! 50! 69!! + 0! d 70!! 0! 3 Simplify the following. a (n + ) n! b (n )(n 2)(n 3)! c n! (n 3)! (n )! (n + )! e n! (n + 2)! 4 Evaluate the following. d (n )! (n + )! f n3 n 2 2n (n + )! (n 2)! n 2 a a 5 2 b b a5 3 b c a2 2 b d 20 C 3 e a 7 0 b f a3 0 b 5 Simplify the following. a a n 3 b b a n n 3 b c an + 3 n b d a 2n + 2n b e an 2 b + an 3 b f an + 3 b 6 Expand the following. a (x + ) 5 b (2 x) 5 c (2x + 3y) 6 d a x 2 + 2b 7 e ax x b 8 f (x 2 + ) 0 78 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 78 7//5 3:2 AM

22 7 Obtain each of the following terms. a The fourth term in the expansion of (5x + 2) 6 b The tenth term in the expansion of ( + 2x) 2 c The sixth term in the expansion of (2x + 3) 0 d The third term in the expansion of (3x 2 ) 6 e The middle term(s) in the expansion of (x 5) 6 f The middle term(s) in the expansion of (x + 2y) 7 8 a Specify the term which contains x 4 in the expansion of (x + 3) 2. b Obtain the coefficient of x 6 in the expansion of ( 2x 2 ) 9. c Express the coefficient of x 5 in the expansion of (3 + 4x) as a product of its prime factors. d Calculate the coefficient of x 2 in the expansion of a x 2 2 x b 8. e Find the term independent of x in the expansion of ax 2 + x 3b 0. f Find the term independent of x in the expansion of ax + x b 6 ax x b 6. 9 a Determine the value of a so that the coefficients of the fourth and the fifth terms in the expansion of ( + ax) 0 are equal. b If the coefficient of x 2 in the expansion of ( + x + x 2 ) 4 is equal to the coefficient of x in the expansion of ( + x) n, find the value of n. 20 a Use the expansion of ( + x) 0 with suitably chosen x to show that 2 0 a 0 0 b + a0 b + a0 b... + a0b and interpret this result for 2 0 Pascal s triangle. b Show that a n + b a n r r b + an b and interpret this result for Pascal s r triangle. Master 2 Evaluate the following using CAS technology. a 5! b a 5 0 b 22 a Solve for n: a n 2 b 770 b Solve for r: a 2 r b 220 I m very well acquainted, too, with matters mathematical; I understand equations, both the simple and quadratical; About Binomial Theorem I am teeming with a lot o news, With many cheerful facts about the square of the hypotenuse! Source: Verse 2 of I am the very model of a modern major general from Pirates of Penzance by Gilbert and Sullivan. Topic 2 Algebraic foundations 79 c02algebraicfoundations.indd 79 7//5 3:2 AM

23 2.5 Unit & 2 AOS 2 Topic Concept 4 Real numbers Concept summary Interactivity Sets int-2556 Sets of real numbers The concept of numbers in counting and the introduction of symbols for numbers marked the beginning of major intellectual development in the minds of the early humans. Every civilisation appears to have developed a system for counting using written or spoken symbols for a few, or more, 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 evolved which collectively form the real number system. Real numbers are all the numbers which are positive or zero or negative. Before further describing and classifying the real number system, a review of some mathematical notation is given. Set notation A set is a collection of objects, these objects being referred to as the elements of the set. The elements may be listed as, for example, the set A 5, 2, 3, 4, 56 and the set B 5, 3, 56. The statement 2 A means 2 is an element of set A, and the statement 2 B means 2 does not belong to, or is not an element of, set B. Since every element in set B 5, 3, 56 is also an element of set A 5, 2, 3, 4, 56, B is a subset of set A. This is written as B A. However we would write A B since A is not a subset of B. The union of the sets A and B contains the elements which are either in A or in B or in both. Element should not be counted twice. The intersection of the sets A and B contains the elements which must be in both A and B. This is written as A B and would be the same as the set B for this example. The exclusion notation A\ B excludes, or removes, any element of B from A. This leaves a set with the elements 52, 46. Sets may be given a description as, for example, set C 5x : < x < 06. The set C is read as C is the set of numbers x such that x is between and 0. The set of numbers not in set C is called the complement of C and given the symbol C. The description of this set could be written as C 5x : x or x 06. A set and its complement cannot intersect. This is written as C C where is a symbol for empty set. Such sets are called disjoint sets. There will be ongoing use of set notation throughout the coming chapters. Classification of numbers While 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, nor 3x 2 where fractions are needed. 80 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 80 7//5 3:2 AM

24 The following sequence of subsets of the real number system, while logical, does not necessarily reflect the historical order in which the real number system was established. For example, fractions were established long before the existence of negative numbers was accepted. Natural numbers are the positive whole numbers or counting numbers. The set of natural numbers is N 5, 2, 3,...6. The positive and negative whole numbers, together with the number zero, are called integers. The set of integers is Z ,, 0,, 2, 3,...6. The symbol Z is derived from the German word zahl for number. Rational numbers are those which can be expressed as quotients in the form p q, where q 0, and p and q are integers which have no common factors other than. The symbol for the set of rational numbers is Q (for quotients). Rational numbers include finite and recurring decimals as well as fractions and integers. For example: , , , and 5 5 are rational. Natural numbers and integers are subsets of the set of rational numbers with N Z Q. Irrational numbers are numbers which are not rational; they cannot be expressed in fraction form as the ratio of two integers. Irrational numbers include numbers such as!2 and π. The set of irrational numbers is denoted by the symbol Q using the complement symbol for not. Q Q as the rational and irrational sets do not intersect. The irrational numbers are further classified into the algebraic irrationals and the non- algebraic ones known as transcendental numbers. Algebraic irrationals are those which, like rational numbers, can be solutions to an equation with integer coefficients, while transcendental numbers cannot. For example, π is transcendental while!2 is algebraic since it is a solution of the equation x R The union of the set of rational and irrational numbers Qʹ forms the set of real numbers R. Hence R Q Q. This is displayed in the diagram showing the subsets of the real numbers. The set of all real numbers forms a number line continuum on which all of the positive or zero or negative numbers are placed. Hence R R 506 R +. Q Z N R Zero R + The sets which formed the building blocks of the real number system have been defined, enabling the real number system to be viewed as the following hierarchy. 0 Real numbers Rationals Irrationals Integers Fractions Algebraic Transcendental Topic 2 Algebraic foundations 8 c02algebraicfoundations.indd 8 7//5 3:2 AM

25 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. Any expression of the form a does not represent a number since division by zero is 0 not possible. If a 0, the expression 0 is said to be indeterminate. It is not defined 0 as a number. To illustrate the second point, consider 3 0. Suppose 3 divided by 0 is possible and results in a number we shall call n. 3 0 n 3 0 n 3 0 The conclusion is nonsensical so 3 is not defined. 0 However, if we try the same process for zero divided by zero, we obtain: 0 0 n 0 0 n 0 0 While the conclusion holds, it is not possible to determine a value for n, so 0 0 is indeterminate. It 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!, 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 3 a Classify each of the following numbers as an element of a subset of the real numbers. i 3 5 ii!7 iii iv!9 b Which of the following are correct statements? i 5 Z ii Z N iii R R + R THINK WRITE a i Fractions are rational numbers. a i 3 5 Q ii Surds are irrational numbers. ii!7 Q 82 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 82 7//5 3:2 AM

26 iii Evaluate the number using the correct order of operations. iii (6 2 3) Z. An alternative answer is (6 2 3) Q. iv Evaluate the square root. iv!9 3!9 N Alternative answers are possible. 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 numbers. ii Z N is incorrect since N Z. iii This is the union of R, the set of negative real numbers, and R +, the set of positive real numbers. iii R R + R is incorrect since R includes the number zero which is neither positive nor negative. Interval notation Interval notation provides an alternative and often convenient way of describing certain sets of numbers. Closed interval [a, b] 5x : a x b6 is the set of real numbers that lie between a and b, including the endpoints, a and b. The inclusion of the endpoints is indicated by the use of the square brackets [ ]. This is illustrated on a number line using closed circles at the endpoints. a b Open interval (a, b) 5x : a < x < b6 is the set of real numbers that lie between a and b, not including the endpoints a and b. The exclusion of the endpoints is indicated by the use of the round brackets ( ). This is illustrated on a number line using open circles at the endpoints. a b Half-open intervals Half-open intervals have only one endpoint included. [a, b) 5x : a x < b6 (a, b] 5x : a < x b6 a b a b Topic 2 Algebraic foundations 83 c02algebraicfoundations.indd 83 7//5 3:2 AM

27 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 4 a Illustrate the following on a number line and express in alternative notation. i ( 2, 2] ii 5x : x 6 iii 5, 2, 3, 46. b Use interval notation to describe the sets of numbers shown on the following number lines. i ii THINK WRITE a i Describe the given interval. Note: The round bracket indicates not included and a square bracket indicates included. a i ( 2, 2] is the interval representing the set of numbers between 2 and 2, closed at 2, open at Write the set in alternative notation. An alternative notation for the set is ( 2, 2] 5x : 2 < x 26. ii Describe the given set. ii 5x : x 6 is the set of all numbers greater than or equal to. This is an infinite interval which has no right-hand endpoint Write the set in alternative notation. An alternative notation is 5x : x 6 [, ). iii Describe the given set. Note: This set does not contain all numbers between the beginning and end of an interval. iii 5, 2, 3, 46 is a set of discrete elements. 2 Write the set in alternative notation. Alternative notations could be 5, 2, 3, 46 5x : x 4, x N6, or 5, 2, 3, 46 [, 4] N. b i Describe the set using interval notation with appropriate brackets. b i The set of numbers lie between 3 and 5, with both endpoints excluded. The set is described as (3, 5). ii Describe the set as the union of the two disjoint intervals. ii The left branch is (, 3] and the right branch is [5, ). The set of numbers is the union of these two. It can be described as (, 3] [5, ). 2 Describe the same set by considering the interval that has been excluded from R. Alternatively, the diagram can be interpreted as showing what remains after the set (3, 5) is excluded from the set R. An alternative description is R\(3, 5). 84 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 84 7//5 3:2 AM

28 Exercise 2.5 PRactise Consolidate Sets of real numbers a Classify each of the following numbers as an element of a subset of the real numbers. ii!27 iii (6 2) 3 iv!0.25 b Which of the following are correct statements? i 7 N ii Q N iii Q Q R x 5 2 For what value(s) of x would be undefined? (x + )(x 3) 3 a Illustrate the following on a number line and express in alternative notation. i [ 2, 2) ii 5x : x < 6 iii 5 2,, 0,, 26 b Use interval notation to describe the sets of numbers shown on the following number lines. i 6 i ii Write R\5x : < x 46 as the union of two sets expressed in interval notation. 5 Which of the following does not represent a real number? a " 3 4 b!0 c d 5π 2 (8 4) e Explain why each of the following statements is false and then rewrite it as a correct statement. a! Q b a 4 9 b Z c R + 5x : x 06 d!2.25 Q 7 Select the irrational numbers from the following set of numbers. 5!, 2,, π,!2, 2 π 6 8 State whether the following are true or false. a R R b N R + c Z N R d Q Z Z e Q Z R \ Q f Z \ N Z 9 Determine any values of x for which the following would be undefined. a b x + 2 x + 8 c x + 5 x 2 (2x + 3)(5 x) 0 Use interval notation to describe the intervals shown on the following number lines. a c b d d 4 x 2 4x Topic 2 Algebraic foundations 85 c02algebraicfoundations.indd 85 7//5 3:2 AM

29 Express the following in interval notation. a 5x : 4 < x 86 b 5x : x > 36 c 5x : x 06 d 5x : 2 x 06 2 Show the following intervals on a number line. a [ 5, 5) b (4, ) c [ 3, 7] d ( 3, 7] e (, 3] f (, ) 3 Illustrate the following on a number line. a R \[ 2, 2] b (,!2) (!2, ) c [ 4, 2) (0, 4) d [ 4, 2) (0, 4) e 5, 0, 6 f R \506 4 Use an alternative form of notation to describe the following sets. a 5x : 2 < x < 6, x Z6 b R \(, 5] c R d (, 4) [2, ) Master 2.6 Unit & 2 AOS 2 Topic Concept 5 Surds Concept summary 5 Determine which of the following are rational and which are irrational numbers. a!7225 b!75600 c The ancient Egyptians devised the formula A 64 8 d2 for calculating the area A of a circle of diameter d. Use this formula to find a rational approximation for π and evaluate it to 9 decimal places. Is it a better approximation than 22 7? Surds A surd is an nth root eg " n x. surds are irrational numbers, and cannot be expressed in the quotient form p. Hence surds have neither a finite nor a recurring decimal form. q Any decimal value obtained from a calculator is just an approximation. All surds have radical signs, such as square roots or cube roots, but not all numbers with radical signs are surds. For surds, the roots cannot be evaluated exactly. Hence,!26 is a surd.!25 is not a surd since 25 is a perfect square,!25 5, which is rational. Ordering surds Surds are real numbers and therefore have a position on the number line. To estimate the position of!6, we can place it between two rational numbers by placing 6 between its closest perfect squares. 86 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 86 7//5 3:2 AM

30 4 < 6 < 9!4 <!6 <!9 2 <!6 < 3 So!6 lies between 2 and 3, closer to 2, since 6 lies closer to 4 than to 9. Checking with a calculator,! Note that the symbol! always gives a positive number, so the negative surd!6 would lie on the number line between 3 and 2 at the approximate position To order the sizes of two surds such as 3!5 and 5!3, express each as an entire surd. 3!5!9!5 5!3!25!3!9 5 and!25 3!45!75 Since!45 <!75 it follows that 3!5 < 5!3. Unit & 2 AOS 2 Topic Concept 6 Simplifying surds Concept summary Surds in simplest form Surds are said to be in simplest form when the number under the square root sign contains no perfect square factors. This means that 3!5 is the simplest form of!45 and 5!3 is the simplest form of!75. If the radical sign is a cube root then the simplest form has no perfect cube factors under the cube root. To express!28 in its simplest form, find perfect square factors of 28.!28!64 2!64!2 8!2!28 8!2 Worked example 5 a Express 56!2, 4!3, 2!5, 76 with its elements in increasing order. b Express in simplest form i!56 ii 2"252a 2 b assuming a > 0. THINK a Express each number entirely as a square root. WRITE a 56!2, 4!3, 2!5, 76 6!2!36!2!72, 4!3!6!3!48, 2!5!4!5!20 and 7!49. 2 Order the terms.!20 <!48 <!49 <!72 umbers is 52!5, 4!3, 7, 6!26. Topic 2 Algebraic foundations 87 c02algebraicfoundations.indd 87 7//5 3:2 AM

31 b i Find a perfect square factor of the number under the square root sign. ii Find any perfect square factors of the number under the square root sign. 2 Express the square root terms as products and simplify where possible. 3 Try to find the largest perfect square factor for greater efficiency. b i!56!4 4!4!4 2!4 ii 2"252a 2 b 2"4 9 7a 2 b 2!4!9!7 "a 2!b 2 2 3!7 a!b 2a!7b Alternatively, recognising that 252 is 36 7, 2"252a 2 b 2"36 7a 2 b 2!36!7 "a 2!b 2 6!7 a!b 2a!7b Operations with surds As surds are real numbers, they obey the usual laws for addition and subtraction of like terms and the laws of multiplication and division. Addition and subtraction a!c + b!c (a + b)!c a!c b!c (a b)!c Surdic expressions such as!2 +!3 cannot be expressed in any simpler form as!2 and!3 are unlike surds. Like surds have the same number under the square root sign. Expressing surds in simplest form enables any like surds to be recognised. Multiplication and division!c!d!(cd) a!c b!d (ab)!(cd)!c!d c Åd Note that!c2 2 c because!c2 2!c!c "c 2 c. 88 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 88 7//5 3:2 AM

32 Worked example 6 Simplify the following. a 3!5 + 7!2 + 6!5 3!2 b 3!98 4!72 + 2!25 c 4!3 6!5 THINK WRITE a Collect like surds together and simplify. a 3!5 + 7!2 + 6!5 3!2 3!5 + 6!5 + 7!2 3!2 9!5 + 4!2 b Write each surd in simplest form. b 3!98 4!72 + 2!25 3!49 2 4! ! !2 4 6! !5 2!2 24!2 + 0!5 2 Collect like surds together. (2!2 24!2) + 0!5 3!2 + 0!5 c Multiply the rational numbers together and multiply the surds together. c 4!3 6!5 (4 6)!(3 5) 24!45 2 Write the surd in its simplest form. 24! !5 72!5 Expansions Expansions of brackets containing surds are carried out using the distributive law in the same way as algebraic expansions.!a(!b +!c)!ab +!ac (!a +!b)(!c +!d)!ac +!ad +!bc +!bd The perfect squares formula for binomial expansions involving surds becomes:!a ±!b2 2 a ± 2!ab + b, since: (!a ±!b) 2 (!a) 2 ± 2!a!b + (!b) 2 a ± 2!ab + b The difference of two squares formula becomes!a +!b2!a!b2 a b since: (!a +!b)(!a!b) (!a) 2 (!b) 2 a b The binomial theorem can be used to expand higher powers of binomial expressions containing surds. Topic 2 Algebraic foundations 89 c02algebraicfoundations.indd 89 7//5 3:2 AM

33 Worked example 7 Expand and simplify the following. a 3!2(4!2 5!6) b (2!3 5!2)(4!5 +!4) THINK c (3!3 + 2!5) 2 d (!7 + 3!2)(!7 3!2) WRITE a Use the distributive law to expand, then simplify each term. a 3!24!2 5!62 2!4 5! ! ! !3 b Expand as for algebraic terms. b 2!3 5!22 4!5 +!42 8!5 + 2!42 20!0 5!28 2 Simplify where possible. 8!5 + 2!42 20!0 5!4 7 8!5 + 2!42 20!0 5 2!7 8!5 + 2!42 20!0 0!7 There are no like surds so no further simplification is possible. c Use the rule for expanding a perfect square. c 3!3 + 2! Simplify each term remembering that!a2 2 a and collect any like terms together. (3!3) 2 + 2(3!3) (2!5) + (2!5) ! ! !5 d Use the rule for expanding a difference of two squares. d!7 + 3!22!7 3!22 (!7) 2 (3!2) Rationalising denominators It is usually desirable to express any fraction whose denominator contains surds as a fraction with a denominator containing only a rational number. This does not necessarily mean the rational denominator form of the fraction is simpler, but it can provide a form which allows for easier manipulation and it can enable like surds to be recognised in a surdic expression. The process of obtaining a rational number on the denominator is called rationalising the denominator. There are different methods for rationalising denominators, depending on how many terms there are in the denominator. 90 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 90 7//5 3:3 AM

34 Rationalising monomial denominators a Consider, where a, b, c Q. This fraction has a monomial denominator since its b!c denominator contains the one term, b!c. In order to rationalise the denominator of this fraction, we use the fact that!c!c c, a rational number. Multiply both the numerator and the denominator by!c. As this is equivalent to multiplying by, the value of the fraction is not altered. a b!c a b!c "c "c a!c b(!c!c) a!c bc a By this process b!c a!c and the denominator, bc, is now rational. bc Once the denominator has been rationalised, it may be possible to simplify the expression if, for example, any common factor exists between the rationals in the numerator and denominator. Rationalising binomial denominators!a +!b and!a!b are called conjugate surds. Multiplying a pair of conjugate surds always results in a rational number since!a +!b2!a!b2 a b. This fact is used to rationalise binomial denominators. Consider where a, b Q. This fraction has a binomial denominator since!a +!b its denominator is the addition of two terms. To rationalise the denominator, multiply both the numerator and the denominator by!a!b, the conjugate of the surd in the denominator. This is equivalent to multiplying by, so the value of the fraction is unaltered; however, it creates a difference of two squares on the denominator. By this process we have rational number.!a +!b "a "b!a +!b "a "b!a!b (!a +!b)(!a!b)!a!b a b!a!b where the denominator, a b, is a!a +!b a b Topic 2 Algebraic foundations 9 c02algebraicfoundations.indd 9 7//5 3:3 AM

35 Worked example 8 a Express the following with a rational denominator. 2 5!3 3!0 i ii 5!3 4!5 b Simplify 5!2 4!2 + 3!8. THINK a 6 c Express with a rational denominator. 5!3 3!2 d Given p 3!2, calculate, expressing the answer with a rational p 2 denominator. i The denominator is monomial. Multiply both numerator and denominator by the surd part of the monomial term. 2 Multiply the numerator terms together and multiply the denominator terms together. 3 Cancel the common factor between the numerator and denominator. ii The denominator is monomial. Multiply both numerator and denominator by the surd part of the monomial term. 2 Simplify the surds, where possible. 3 Take out the common factor in the numerator since it can be cancelled as a factor of the denominator. WRITE a i 2 5"3 2 5!3!3!3 2! ! ! !3 5 ii 5!3 3!0 4!5 5!3 3!02!5 4!5!5!55!3 3! !5 3! !5 3 5!2 20 5!5 5!2 20 5!5 3!22 20!5 3! Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 92 7//5 3:3 AM

36 b Rationalise any denominators containing surds and simplify all terms in order to identify any like surds that can be collected together. c The denominator is binomial. Multiply both numerator and denominator by the conjugate of the binomial surd contained in the denominator. 2 Expand the difference of two squares in the denominator. Note: This expansion should always result in a rational number. b 5!2 4!2 + 3!8 5!2 4!2!2! !2 5!2 4!2 + 9!2 2 5!2 2!2 + 9!2 2!2 c The conjugate of 5!3 3!2 is 5!3 + 3!2 6 5!3 3!2 6 5!3 + 3!2 5!3 3!2 5!3 + 3!2 65!3 + 3!22 5!3 3!22 5!3 + 3!22 65!3 + 3!22 5!32 2 3! !3 + 3! !3 + 3! Cancel the common factor between the numerator and the denominator. Note: The numerator could be expanded but there is no further simplification to gain by doing so. 2 6(5!3 + 3!2) (5!3 + 3!2) or 9 d Substitute the given value and simplify. d Given p 3!2, p 2 (3!2 ) 2 (9 2 6!2 + ) 8 6!2 2 Factorise the denominator so that the binomial surd is simpler. 63!22 0!3 + 6!2 9 Topic 2 Algebraic foundations 93 c02algebraicfoundations.indd 93 7//5 3:3 AM

37 3 Multiply numerator and denominator by the conjugate of the binomial surd contained in the denominator. 4 Expand the difference of two squares and simplify. 63!22 3 +!2 3 +!2 3 +!2 63!22 3 +!22 3 +!2 6 (3) 2! !2 6(9 2) 3 +! !2 42 Exercise 2.6 PRactise Surds WE5 a Express 53!3, 4!5, 5!2, 56 with its elements in increasing order. b Express the following in simplest form. i!84 ii 2"08ab 2 assuming b > 0 2 Express " in simplest form. 3 WE6 Simplify the following. a!5 4!7 7!5 + 3!7 b 3!48 4!27 + 3!32 c 3!5 7!5 4 Simplify 2!2!0!80 + +! !2 5 WE7 Expand and simplify the following. a 2!34!5 + 5!32 b!3 8!22 5!5 2!22 c 4!3 5!22 2 d 3!5 2!2 3!5 + 2!2 6 a Expand! b If!2 +!62 2 2!3!2 +!62!2!62 a + b!3, a, b N, find the values of a and b. 7 WE8 a Express the following with a rational denominator. 6 3!5 + 7!5 i ii 7"2 2!3 b Simplify 4!6 + 2!6 5!24. 0 c Express with a rational denominator. 4!3 + 3!2 d Given p 4!3 +, calculate, expressing the answer with a rational p 2 denominator. 94 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 94 7//5 3:3 AM

38 Consolidate 8 a Simplify 2!3!3 +!3 by first rationalising each denominator.!3 + 2 b Show that!3!3 + +!3 + is rational by first placing each fraction on a!3 common denominator. 9 Select the surds from the following set of real numbers. 4 e!8,!900, Å 9,!.44, "03, π, " 3 27, " 3 36 f 0 Express the following as entire surds. a 4!5 b 2" 3 6 c 9!7 4 3 d!3 e ab!c f m" 3 n Express each of the following in simplified form. a!75 b 5!48 c!2000 d 3!288 e 2!72 f " Simplify the following. a 3!7 + 8!3 + 2!7 9!3 b 0!2 2!6 + 4!6 8!2 c 3!50!8 d 8!45 + 2!25 e!6 + 7!5 + 4!24 8!20 f 2!2 7! !8 2 3!62 3 Carry out the following operations and express answers in simplest form. a 4!5 2!7 b 0!6 8!0 c 3!8 2!5 d!8!72 e 4!27!47 2!3 f 5!2!3 4!5!6 6 4 Expand and simplify the following. a!23!5 7!62 b 5!37 3!3 + 2!62 c 2!0 3!63!5 + 2!62 d 2!3 +!52 3!2 + 4!72 + 3!2 7!0 e 5!2 3!62 2!3 + 3!02 f " 3 x " 3 y2 " 3 x2 2 + " 3 xy + " 3 y Expand the following. a 2! b 3!6 2!32 2 c!7!52 3 d 2!5 +!32 2!5!32 e 0!2 3!52 0!2 + 3!52 f!3 +!2 + 2!3 +!2 2 6 Express the following in simplest form with rational denominators. a 3"2 4"3 d!6 +!2!5 +!2 b!2 e 2!0 + 5!0 c f!2 3!2 2!8 3!3 + 2!2!3 +!2 Topic 2 Algebraic foundations 95 c02algebraicfoundations.indd 95 7//5 3:3 AM

39 Master 7 Express the following as a single fraction in simplest form. a 4!5 2!6 + 3!6 0 b!2(2!0 + 9!8)!5 3!5!5 2 3 c 2!3(!2 +!3) + 2 2!3!2 2!3 +!2 d + 2!3 2!3 +!2 2!3!2 e!2!4 + 2!2 6 4!7 9!7 f (2!3)2 2 +!3 + 2!3 4 3!2 8 a If x 2!3!0, calculate the value of the following. i x + x ii x 2 4!3x b If y!7 + 2, calculate the value of the following.!7 2 i y ii y y 2 c Determine the values of m and n for which each of the following is a correct statement. i!7 +!3 m!7 + n!3!7!3 ii (2 +!3) 4 7!3 (2 +!3) m +!n 2 d The real numbers x and x 2 are a pair of conjugates. If x b + "b2 4ac : 2a i state x 2 ii calculate the sum x + x 2 iii calculate the product x x 2. 9 A triangle has vertices at the points A(!2, ), B(!5,!0) and C(!0,!5). a Calculate the lengths of each side of the triangle in simplest surd form. b An approximation attributable to the Babylonians is that "a 2 ± b a ± b 2a. Use this formula to calculate approximate values for the lengths of each side of the triangle. c Calculate from the surd form the length of the longest side to decimal place. 20 A rectangular lawn has dimensions!6 +!3 + 2 m by! m. Hew agrees to mow the lawn for the householder. a Calculate the exact area of the lawn. b If the householder received change of $23.35 from $50, what was the cost per square metre that Hew charged for mowing the lawn? Aristotle was probably the first to prove " 2 was what we call irrational and what he called incommensurable. Plato, an ancient Greek philosopher, claimed his teacher Theodorus of Cyrene, building on Aristotle s approach, was the first to prove the irrationality of the non perfect squares from 3 to 7. The work of Theodorus no longer exists. 96 Maths Quest MATHEMATICAL METHODS VCE Units and 2 c02algebraicfoundations.indd 96 7//5 3:3 AM

40 ONLINE ONLY 2.7 Review The Maths Quest Review is available in a customisable format for you to demonstrate your knowledge of this topic. The Review contains: short-answer questions providing you with the opportunity to demonstrate the skills you have developed to efficiently answer questions without the use of CAS technology Multiple-choice questions providing you with the opportunity to practise answering questions using CAS technology ONLINE ONLY Activities To access ebookplus activities, log on to Interactivities A comprehensive set of relevant interactivities to bring difficult mathematical concepts to life can be found in the Resources section of your ebookplus. Extended-response questions providing you with the opportunity to practise exam-style questions. a summary of the key points covered in this topic is also available as a digital document. REVIEW QUESTIONS Download the Review questions document from the links found in the Resources section of your ebookplus. studyon is an interactive and highly visual online tool that helps you to clearly identify strengths and weaknesses prior to your exams. You can then confidently target areas of greatest need, enabling you to achieve your best results. Units & 2 Algebraic foundations Sit topic test Topic 2 AlgebrAic foundations c02algebraicfoundations.indd //5 3:3 AM