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Algebra and
Surds
TERMINOLOGY
Binomial: A mathematical expression consisting of
two terms such as x + 3 or 3x - 1
Binomial product: The product of two binomial
expressions such as (x + 3) (2x - 4)
Expression: A mathematical statement involving numbers,
pronumerals and symbols e.g. 2x - 3
Factorise: The process of writing an expression as a
product of its factors. It is the reverse operation of
expanding brackets i.e. take out the highest common
factor in an expression and place the rest in brackets
e.g. 2y - 8 = 2 (y - 4)
Pronumeral: A letter or symbol that stands for a number
Rationalising the denominator: A process for replacing a
surd in the denominator by a rational number without
altering its value
Surd: From ‘absurd’. The root of a number that has an
irrational value e.g. 3 . It cannot be expressed as a
rational number
Term: An element of an expression containing
pronumerals and/or numbers separated by an operation
such as + , - , # or ' e.g. 2x, - 3
Trinomial: An expression with three terms such as
3x 2 - 2x + 1
Chapter 2 Algebra and Surds
45
INTRODUCTION
THIS CHAPTER REVIEWS ALGEBRA skills, including simplifying expressions,
removing grouping symbols, factorising, completing the square and
simplifying algebraic fractions. Operations with surds, including rationalising
the denominator, are also studied in this chapter.
DID YOU KNOW?
One of the earliest mathematicians to use algebra was Diophantus of Alexandria. It is not known
when he lived, but it is thought this may have been around 250 AD.
In Baghdad around 700–800 AD a mathematician named Mohammed Un-Musa
Al-Khowarezmi wrote books on algebra and Hindu numerals. One of his books was named
Al-Jabr wa’l Migabaloh, and the word algebra comes from the first word in this title.
Simplifying Expressions
Addition and subtraction
EXAMPLES
Simplify
1. 7x - x
Solution
Here x is called a
pronumeral.
7x - x = 7x - 1 x
= 6x
2. 4x 2 - 3x 2 + 6x 2
Solution
4x 2 - 3x 2 + 6x 2 = x 2 + 6 x 2
= 7x 2
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
3. x 3 - 3x - 5x + 4
Only add or subtract ‘like’
terms. These have the
same pronumeral (for
example, 3x and 5x).
Solution
x 3 - 3 x - 5x + 4 = x 3 - 8 x + 4
4. 3a - 4b - 5a - b
Solution
3a - 4b - 5a - b = 3a - 5a - 4b - b
= - 2a - 5b
2.1 Exercises
Simplify
1.
2x + 5x
16. 7b + b - 3b
2.
9a - 6a
17. 3b - 5b + 4b + 9b
3.
5z - 4z
18. - 5x + 3x - x - 7x
4.
5a + a
19. 6x - 5y - y
5.
4b - b
20. 8a + b - 4b - 7a
6.
2r - 5r
21. xy + 2y + 3xy
7.
- 4y + 3y
22. 2ab 2 - 5ab 2 - 3ab 2
8.
- 2x - 3x
23. m 2 - 5m - m + 12
9.
2a - 2a
24. p 2 - 7p + 5p - 6
10. - 4k + 7k
25. 3x + 7y + 5x - 4y
11. 3t + 4t + 2t
26. ab + 2b - 3ab + 8b
12. 8w - w + 3w
27. ab + bc - ab - ac + bc
13. 4m - 3m - 2m
28. a 5 - 7x 3 + a 5 - 2x 3 + 1
14. x + 3x - 5x
29. x 3 - 3xy 2 + 4x 2 y - x 2 y + xy 2 + 2y 3
15. 8h - h - 7h
30. 3x 3 - 4x 2 - 3x + 5x 2 - 4x - 6
Chapter 2 Algebra and Surds
47
Multiplication
EXAMPLES
Simplify
1. - 5x # 3y # 2x
Solution
- 5x # 3y # 2x = - 30xyx
= - 30x 2 y
2. - 3x 3 y 2 # - 4xy 5
Solution
Use index laws
to simplify this
question.
- 3x 3 y 2 # - 4xy 5 = 12x 4 y 7
2.2 Exercises
Simplify
1.
5 # 2b
5
11. ^ 2x 2h
2.
2x # 4y
12. 2ab 3 # 3a
3.
5p # 2p
13. 5a 2 b # - 2ab
4.
- 3z # 2w
14. 7pq 2 # 3p 2 q 2
5.
- 5a # - 3b
15. 5ab # a 2 b 2
6.
x # 2y # 7z
16. 4h 3 # - 2h 7
7.
8ab # 6c
17. k 3 p # p 2
8.
4d # 3d
4
18. ^ - 3t 3 h
9.
3a # 4a # a
19. 7m 6 # - 2m 5
10. ^ - 3y h3
20. - 2x 2 # 3x 3 y # - 4xy 2
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Maths In Focus Mathematics Extension 1 Preliminary Course
Division
Use cancelling or index laws to simplify divisions.
EXAMPLES
Simplify
1. 6v 2 y ' 2vy
Solution
By cancelling,
6v 2 y ' 2vy =
=
6v 2 y
2vy
63 # v # v1 # y1
21 # v # y1
= 3v
Using index laws,
6v 2 y ' 2vy = 3v 2 - 1 y 1 - 1
= 3v 1 y 0
= 3v
2.
5a 3 b
15ab 2
Solution
5a 3 b
= 1 a3 -1 b1- 2
3
15ab 2
= 1 a 2 b -1
3
a2
=
3b
2.3 Exercises
Simplify
1.
30x ' 5
2.
2y ' y
3.
4.
5.
8a
2
6.
xy
2x
7.
12p 3 ' 4p 2
8.
3a 2 b 2
6ab
9.
20x
15xy
10.
- 9x 7
3x 4
2
8a 2
a
8a 2
2a
Chapter 2 Algebra and Surds
11. -15ab ' - 5b
12.
2ab
6a 2 b 3
13.
- 8p
4pqs
16.
7pq 3
17. 5a 9 b 4 c - 2 ' 20a 5 b -3 c -1
2 ^ a -5 h b 4
2
18.
14. 14cd 2 ' 21c 3 d 3
15.
42p 5 q 4
-1
4a - 9 ^ b 2 h
19. - 5x 4 y 7 z ' 15xy 8 z - 2
2xy 2 z 3
20. - 9 ^ a 4 b -1 h ' -18a -1 b 3
3
4x 3 y 2 z
Removing grouping symbols
The distributive law of numbers is given by
a ] b + c g = ab + ac
EXAMPLE
7 # (9 + 11) = 7 # 20
= 140
Using the distributive law,
7 # (9 + 11) = 7 # 9 + 7 # 11
= 63 + 77
= 140
This rule is used in algebra to help remove grouping symbols.
EXAMPLES
Expand and simplify.
1. 2 ] a + 3 g
Solution
2 (a + 3) = 2 # a + 2 # 3
= 2a + 6
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
2. - ] 2x - 5 g
Solution
-(2x - 5) = -1 (2x - 5)
= -1 # 2x - 1 # - 5
= - 2x + 5
3. 5a 2]4 + 3ab - c g
Solution
5a 2 (4 + 3ab - c) = 5a 2 # 4 + 5a 2 # 3ab - 5a 2 # c
= 20a 2 + 15a 3 b - 5a 2 c
4. 5 - 2 ^ y + 3 h
Solution
5 - 2 (y + 3 ) = 5 - 2 # y - 2 # 3
= 5 - 2y - 6
= - 2y - 1
5. 2 ] b - 5 g - ] b + 1 g
Solution
2 (b - 5) - (b + 1) = 2 # b + 2 # - 5 - 1 # b -1 # 1
= 2b - 10 - b - 1
= b - 11
2.4 Exercises
Expand and simplify
1.
2]x - 4 g
7.
ab ] 2a + b g
2.
3 ] 2h + 3 g
8.
5n ] n - 4 g
3.
-5 ] a - 2 g
9.
3x 2 y _ xy + 2y 2 i
4.
x ^ 2y + 3 h
10. 3 + 4 ] k + 1 g
5.
x]x - 2 g
11. 2 ] t - 7 g - 3
6.
2a ] 3a - 8 b g
12. y ^ 4y + 3 h + 8y
Chapter 2 Algebra and Surds
13. 9 - 5 ] b + 3 g
20. 2ab ] 3 - a g - b ] 4a - 1 g
14. 3 - ] 2x - 5 g
21. 5x - ] x - 2 g - 3
15. 5] 3 - 2m g + 7 ] m - 2 g
22. 8 - 4 ^ 2y + 1 h + y
16. 2 ] h + 4 g + 3 ] 2h - 9 g
23. ] a + b g - ] a - b g
17. 3 ] 2d - 3 g - ] 5d - 3 g
24. 2 ] 3t - 4 g - ] t + 1 g + 3
18. a ] 2a + 1 g - ^ a 2 + 3a - 4 h
25. 4 + 3 ] a + 5 g - ] a - 7 g
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19. x ] 3x - 4 g - 5 ] x + 1 g
Binomial Products
A binomial expression consists of two numbers, for example x + 3.
A set of two binomial expressions multiplied together is called a binomial
product.
Example: ] x + 3 g ] x - 2 g.
Each term in the first bracket is multiplied by each term in the second
bracket.
] a + b g ^ x + y h = ax + ay + bx + by
Proof
]a + bg]c + d g = a ]c + d g + b ]c + d g
= ac + ad + bc + bd
EXAMPLES
Expand and simplify
1. ^ p + 3h^ q - 4h
Solution
^ p + 3 h ^ q - 4 h = pq - 4p + 3q - 12
2. ]a + 5g2
Solution
] a + 5 g2 = (a + 5)(a + 5)
= a 2 + 5a + 5a + 25
= a 2 + 10a + 25
Can you see a quick
way of doing this?
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Maths In Focus Mathematics Extension 1 Preliminary Course
The rule below is not a binomial product (one expression is a trinomial), but it
works the same way.
] a + b g ^ x + y + z h = ax + ay + az + bx + by + bz
EXAMPLE
Expand and simplify ] x + 4 g ^ 2x - 3y - 1 h .
Solution
(x + 4) (2x - 3y - 1) = 2x 2 - 3xy - x + 8x - 12y - 4
= 2x 2 - 3xy + 7x - 12y - 4
2.5 Exercises
Expand and simplify
1.
]a + 5g]a + 2g
17. ]a + 2bg]a - 2bg
2.
]x + 3g]x - 1g
18. ^ 3x - 4y h^ 3x + 4y h
3.
^ 2y - 3h^ y + 5h
19. ]x + 3g]x - 3g
4.
]m - 4g]m - 2g
20. ^ y - 6h^ y + 6h
5.
]x + 4g]x + 3g
21. ] 3a + 1 g ] 3a - 1 g
6.
^ y + 2h^ y - 5h
22. ]2z - 7g]2z + 7g
7.
]2x - 3g]x + 2g
23. ]x + 9g^ x - 2y + 2h
8.
]h - 7g]h - 3g
24. ] b - 3 g ] 2a + 2b - 1 g
9.
]x + 5g]x - 5g
25. ]x + 2g^ x 2 - 2x + 4h
10. ] 5a - 4 g ] 3a - 1 g
26. ]a - 3g^ a 2 + 3a + 9h
11. ^ 2y + 3h^ 4y - 3h
27. ]a + 9g2
12. ]x - 4g^ y + 7h
28. ]k - 4g2
13. ^ x 2 + 3h]x - 2g
29. ]x + 2g2
14. ]n + 2g]n - 2g
30. ^ y - 7h2
15. ]2x + 3g]2x - 3g
31. ]2x + 3g2
16. ^ 4 - 7y h^ 4 + 7y h
32. ]2t - 1g2
Chapter 2 Algebra and Surds
33. ]3a + 4bg2
37. ] a + b g2
34. ^ x - 5y h2
38. ] a - b g2
35. ]2a + bg2
39. ] a + b g ^ a 2 - ab + b 2 h
36. ] a - b g ] a + b g
40. ] a - b g ^ a 2 + ab + b 2 h
Some binomial products have special results and can be simplified quickly
using their special properties. Binomial products involving perfect squares
and the difference of two squares occur in many topics in mathematics. Their
expansions are given below.
Difference of 2 squares
] a + b g ] a - b g = a2 - b2
Proof
(a + b) (a - b) = a 2 - ab + ab - b 2
= a2 - b2
Perfect squares
] a + b g2 = a 2 + 2ab + b 2
Proof
] a + b g2 = (a + b) (a + b)
= a 2 + ab + ab + b 2
= a 2 + 2ab + b 2
]a - bg2 = a 2 - 2ab + b 2
Proof
] a - b g2 = (a - b) (a - b)
= a 2 - ab - ab + b 2
= a 2 - 2ab + b 2
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
Expand and simplify
1. ]2x - 3g2
Solution
] 2x - 3 g2 = ] 2x g2 - 2 (2x) 3 + 3 2
= 4x 2 - 12x + 9
2. ^ 3y - 4h^ 3y + 4h
Solution
(3y - 4) (3y + 4) = ^ 3y h2 - 4 2
= 9y 2 - 16
2.6 Exercises
Expand and simplify
1.
]t + 4g2
16. ^ p + 1 h ^ p - 1 h
2.
]z - 6g2
17. ]r + 6g]r - 6g
3.
] x - 1 g2
18. ] x - 10 g ] x + 10 g
4.
^ y + 8h2
19. ]2a + 3g]2a - 3g
5.
^ q + 3h2
20. ^ x - 5y h^ x + 5y h
6.
]k - 7g2
21. ] 4a + 1 g ] 4a - 1 g
7.
] n + 1 g2
22. ]7 - 3xg]7 + 3xg
8.
]2b + 5g2
23. ^ x 2 + 2h^ x 2 - 2h
9.
]3 - xg2
2
24. ^ x 2 + 5h
10. ^ 3y - 1 h2
25. ]3ab - 4cg]3ab + 4c g
11. ^ x + y h2
2 2
26. b x + x l
12. ] 3a - b g2
13. ]4d + 5eg2
1
1
27. b a - a lb a + a l
14. ]t + 4g]t - 4g
28. _ x + 6 y - 2 @ i _ x - 6 y - 2 @ i
15. ] x - 3 g ] x + 3 g
29. 6]a + bg + c @2
Chapter 2 Algebra and Surds
30. 7 ] x + 1 g - y A
36. ] x - 4 g3
2
55
Expand (x - 4) (x - 4) 2 .
31. ] a + 3 g2 - ] a - 3 g2
1 2
1 2
37. b x - x l - b x l + 2
32. 16 - ]z - 4g]z + 4g
38. _ x 2 + y 2 i - 4x 2 y 2
33. 2x + ]3x + 1g2 - 4
39. ]2a + 5g3
34. ^ x + y h2 - x ^ 2 - y h
40. ] 2x - 1 g ] 2x + 1 g ] x + 2 g2
2
35. ] 4n - 3 g ] 4n + 3 g - 2n 2 + 5
PROBLEM
Find values of all pronumerals that make this true.
a b
d
f e
i i i h
i i c c
c
e
b
g
b
#
Try c = 9.
Factorisation
Simple factors
Factors are numbers that exactly divide or go into an equal or larger number,
without leaving a remainder.
EXAMPLES
The numbers 1, 2, 3, 4, 6, 8, 12 and 24 are all the factors of 24.
Factors of 5x are 1, 5, x and 5x.
To factorise an expression, we use the distributive law.
ax + bx = x ] a + b g
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
Factorise
1. 3x + 12
Solution
Divide each term by 3 to
find the terms inside the
brackets.
The highest common factor is 3.
3x + 12 = 3 ] x + 4 g
2. y 2 - 2y
Solution
Check answers by
expanding brackets.
The highest common factor is y.
y 2 - 2y = y ^ y - 2 h
3. x 3 - 2x 2
Solution
x and x2 are both common factors. We take out the highest common
factor which is x2.
x 3 - 2x 2 = x 2 ] x - 2 g
4. 5] x + 3 g + 2y ] x + 3 g
Solution
The highest common factor is x + 3.
5 ] x + 3 g + 2y ] x + 3 g = ] x + 3 g ^ 5 + 2 y h
5. 8a 3 b 2 - 2ab 3
Solution
There are several common factors here. The highest common
factor is 2ab2.
8a 3 b 2 - 2ab 3 = 2ab 2 ^ 4a 2 - bh
Chapter 2 Algebra and Surds
2.7 Exercises
Factorise
1.
2y + 6
19. x ] m + 5 g + 7 ] m + 5 g
2.
5x - 10
20. 2 ^ y - 1 h - y ^ y - 1 h
3.
3m - 9
21. 4^ 7 + y h - 3x ^ 7 + y h
4.
8x + 2
22. 6x ]a - 2g + 5]a - 2g
5.
24 - 18y
23. x ] 2t + 1 g - y ] 2t + 1 g
6.
x 2 + 2x
7.
m 2 - 3m
24. a ] 3x - 2 g + 2b ] 3x - 2 g
- 3c ] 3x - 2 g
8.
2y 2 + 4y
9.
15a - 3a 2
25. 6x 3 + 9x 2
26. 3pq 5 - 6q 3
27. 15a 4 b 3 + 3ab
10. ab 2 + ab
28. 4x 3 - 24x 2
11. 4x 2 y - 2xy
29. 35m 3 n 4 - 25m 2 n
12. 3mn 3 + 9mn
30. 24a 2 b 5 + 16ab 2
13. 8x 2 z - 2xz 2
14. 6ab + 3a - 2a
31. 2rr 2 + 2rrh
2
32. ]x - 3g2 + 5]x - 3g
15. 5x 2 - 2x + xy
33. y 2 ]x + 4g + 2]x + 4g
16. 3q 5 - 2q 2
34. a ] a + 1 g - ] a + 1 g2
17. 5b 3 + 15b 2
35. 4ab ^ a 2 + 1 h - 3 ^ a 2 + 1 h
18. 6a 2 b 3 - 3a 3 b 2
Grouping in pairs
If an expression has 4 terms, it may be factorised in pairs.
ax + bx + ay + by = x(a + b) + y (a + b)
= ( a + b) ( x + y)
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
Factorise
1. x 2 - 2x + 3x - 6
Solution
x 2 - 2x + 3x - 6 = x (x - 2) + 3 (x - 2)
= (x - 2) (x + 3)
2. 2x - 4 + 6y - 3xy
Solution
2x - 4 + 6y - 3xy = 2 (x - 2) + 3y (2 - x)
= 2 ( x - 2) - 3y ( x - 2 )
= (x - 2) (2 - 3y)
or 2x - 4 + 6y - 3xy = 2 (x - 2) - 3y (- 2 + x)
= 2 ( x - 2) - 3y ( x - 2 )
= (x - 2) (2 - 3y)
2.8 Exercises
Factorise
1.
2x + 8 + bx + 4b
12. m - 2 + 4y - 2my
2.
ay - 3a + by - 3b
13. 2x 2 + 10xy - 3xy - 15y 2
3.
x 2 + 5x + 2x + 10
14. a 2 b + ab 3 - 4a - 4b 2
4.
m 2 - 2m + 3m - 6
15. 5x - x 2 - 3x + 15
5.
ad - ac + bd - bc
16. x 4 + 7x 3 - 4x - 28
6.
x 3 + x 2 + 3x + 3
17. 7x - 21 - xy + 3y
7.
5ab - 3b + 10a - 6
18. 4d + 12 - de - 3e
8.
2xy - x 2 + 2y 2 - xy
19. 3x - 12 + xy - 4y
9.
ay + a + y + 1
20. 2a + 6 - ab - 3b
10. x 2 + 5x - x - 5
21. x 3 - 3x 2 + 6x - 18
11. y + 3 + ay + 3a
22. pq - 3p + q 2 - 3q
Chapter 2 Algebra and Surds
23. 3x 3 - 6x 2 - 5x + 10
27. 4x 3 - 6x 2 + 8x - 12
24. 4a - 12b + ac - 3bc
28. 3a 2 + 9a + 6ab + 18b
25. xy + 7x - 4y - 28
29. 5y - 15 + 10xy - 30x
26. x 4 - 4x 3 - 5x + 20
30. rr 2 + 2rr - 3r - 6
59
Trinomials
A trinomial is an expression with three terms, for example x 2 - 4x + 3.
Factorising a trinomial usually gives a binomial product.
x 2 + ] a + b g x + ab = ] x + a g ] x + b g
Proof
x 2 + (a + b) x + ab = x 2 + ax + bx + ab
= x(x + a) + b(x + a)
= (x + a) (x + b)
EXAMPLES
Factorise
1. m 2 - 5m + 6
Solution
a + b = - 5 and ab = + 6
-2
+6 '
-3
-5
Numbers with sum - 5 and product + 6 are - 2 and - 3.
` m 2 - 5m + 6 = [m + ] - 2 g] [m + ] - 3 g]
= ]m - 2g]m - 3g
Guess and check by
trying - 2 and - 3
or -1 and - 6.
2. y 2 + y - 2
Solution
a + b = + 1 and ab = - 2
+2
-2 '
-1
+1
Two numbers with sum + 1 and product - 2 are + 2 and -1.
` y2 + y - 2 = ^ y + 2 h ^ y - 1 h
Guess and check by
trying 2 and -1 or
- 2 and 1.
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Maths In Focus Mathematics Extension 1 Preliminary Course
2.9 Exercises
Factorise
1.
x 2 + 4x + 3
14. a 2 - 4a + 4
2.
y 2 + 7y + 12
15. x 2 + 14x - 32
3.
m 2 + 2m + 1
16. y 2 - 5y - 36
4.
t 2 + 8t + 16
17. n 2 - 10n + 24
5.
z2 + z - 6
18. x 2 - 10x + 25
6.
x 2 - 5x - 6
19. p 2 + 8p - 9
7.
v 2 - 8v + 15
20. k 2 - 7k + 10
8.
t 2 - 6t + 9
21. x 2 + x - 12
9.
x 2 + 9x - 10
22. m 2 - 6m - 7
10. y 2 - 10y + 21
23. q 2 + 12q + 20
11. m 2 - 9m + 18
24. d 2 - 4d - 5
12. y 2 + 9y - 36
25. l 2 - 11l + 18
13. x 2 - 5x - 24
The result x 2 + ] a + b g x + ab = ] x + a g ] x + b g only works when the coefficient
of x 2 (the number in front of x 2) is 1. When the coefficient of x 2 is not 1, for
example in the expression 5x 2 - 2x + 4, we need to use a different method to
factorise the trinomial.
There are different ways of factorising these trinomials. One method is
the cross method. Another is called the PSF method. Or you can simply guess
and check.
EXAMPLES
Factorise
1. 5y 2 - 13y + 6
Solution—guess and check
For 5y2, one bracket will have 5y and the other y:
^ 5y h ^ y h .
Now look at the constant (term without y in it): + 6.
Chapter 2 Algebra and Surds
The two numbers inside the brackets must multiply to give + 6.
To get a positive answer, they must both have the same signs.
But there is a negative sign in front of 13y so the numbers cannot be both
positive. They must both be negative.
^ 5y - h ^ y - h
To get a product of 6, the numbers must be 2 and 3 or 1 and 6.
Guess 2 and 3 and check:
^ 5y - 2 h ^ y - 3 h = 5y 2 - 15y - 2y + 6
= 5y 2 - 17y + 6
This is not correct.
Notice that we are mainly interested in checking the middle two terms,
-15y and - 2y.
Try 2 and 3 the other way around:
^ 5y - 3 h ^ y - 2 h .
Checking the middle terms: -10y - 3y = -13y
This is correct, so the answer is ^ 5y - 3 h ^ y - 2 h .
Note: If this did not check out, do the same with 1 and 6.
Solution—cross method
Factors of 5y 2 are 5y and y.
Factors of 6 are -1 and - 6 or - 2 and - 3.
Possible combinations that give a middle term of -13y are
5y
-2
5y
-3
5y
-1
5y
-6
y
-3
y
-2
y
-6
y
-1
By guessing and checking, we choose the correct combination.
-3
5y # - 2 = -10y
5y
y
-2
y # - 3 = - 3y
-13y
` 5y 2 - 13y + 6 = ^ 5y - 3 h ^ y - 2 h
Solution—PSF method
P: Product of first and last terms
S: Sum or middle term
F: Factors of P that give S
- 3y
30y 2 )
-10y
-13y
30y 2
-13y
- 3y, -10y
` 5y 2 - 13y + 6 = 5y 2 - 3y - 10y + 6
= y ^ 5y - 3 h - 2 ^ 5 y - 3 h
= ^ 5y - 3 h ^ y - 2 h
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
2. 4y 2 + 4y - 3
Solution—guess and check
For 4y2, both brackets will have 2y or one bracket will have 4y and the
other y.
Try 2y in each bracket:
^ 2y h ^ 2y h .
Now look at the constant: - 3.
The two numbers inside the brackets must multiply to give - 3.
To get a negative answer, they must have different signs.
^ 2y - h ^ 2y + h
To get a product of 3, the numbers must be 1 and 3.
Guess and check:
^ 2y - 3 h ^ 2 y + 1 h
Checking the middle terms: 2y - 6y = - 4y
This is almost correct, as the sign is wrong but the coefficient is right
(the number in front of y).
Swap the signs around:
^ 2y - 1 h ^ 2 y + 3 h = 4y 2 + 6 y - 2 y - 3
= 4y 2 + 4y - 3
This is correct, so the answer is ^ 2y - 1 h ^ 2y + 3 h .
Solution—cross method
Factors of 4y 2 are 4y and y or 2y and 2y.
Factors of 3 are -1 and 3 or - 3 and 1.
Trying combinations of these factors gives
3
2y
2y # - 1 = - 2 y
2y
-1
2y # 3 =
6y
4y
` 4y 2 + 4y - 3 = ^ 2 y + 3 h ^ 2 y - 1 h
Solution—PSF method
P: Product of first and last terms
-12y 2
S: Sum or middle term
4y
F: Factors of P that give S
+ 6y, - 2y
2 + 6y
-12y )
-2y
+ 4y
` 4y 2 + 4y - 3 = 4 y 2 + 6 y - 2 y - 3
= 2y ^ 2y + 3 h - 1 ^ 2 y + 3 h
= ^ 2y + 3 h ^ 2y - 1 h
Chapter 2 Algebra and Surds
2.10
Exercises
Factorise
1.
2a 2 + 11a + 5
16. 4n 2 - 11n + 6
2.
5y 2 + 7y + 2
17. 8t 2 + 18t - 5
3.
3x 2 + 10x + 7
18. 12q 2 + 23q + 10
4.
3x 2 + 8x + 4
19. 8r 2 + 22r - 6
5.
2b 2 - 5b + 3
20. 4x 2 - 4x - 15
6.
7x 2 - 9x + 2
21. 6y 2 - 13y + 2
7.
3y 2 + 5y - 2
22. 6p 2 - 5p - 6
8.
2x 2 + 11x + 12
23. 8x 2 + 31x + 21
9.
5p 2 + 13p - 6
24. 12b 2 - 43b + 36
10. 6x 2 + 13x + 5
25. 6x 2 - 53x - 9
11. 2y 2 - 11y - 6
26. 9x 2 + 30x + 25
12. 10x 2 + 3x - 1
27. 16y 2 + 24y + 9
13. 8t 2 - 14t + 3
28. 25k 2 - 20k + 4
14. 6x 2 - x - 12
29. 36a 2 - 12a + 1
15. 6y 2 + 47y - 8
30. 49m 2 + 84m + 36
Perfect squares
You have looked at some special binomial products, including
]a + bg2 = a 2 + 2ab + b 2 and ]a - bg2 = a 2 - 2ab + b 2 .
When factorising, use these results the other way around.
a 2 + 2ab + b 2 = ] a + b g2
a 2 - 2ab + b 2 = ] a - b g2
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
In a perfect square, the
constant term is always a
square number.
Factorise
1. x 2 - 8x + 16
Solution
x 2 - 8x + 16 = x 2 - 2 (4) x + 4 2
= ] x - 4 g2
2. 4a 2 + 20a + 25
Solution
4a 2 + 20a + 25 = ] 2a g2 + 2 (2a) (5) + 5 2
= ] 2a + 5 g2
2.11
Exercises
Factorise
1.
y 2 - 2y + 1
12. 16k 2 - 24k + 9
2.
x 2 + 6x + 9
13. 25x 2 + 10x + 1
3.
m 2 + 10m + 25
14. 81a 2 - 36a + 4
4.
t 2 - 4t + 4
15. 49m 2 + 84m + 36
5.
x 2 - 12x + 36
16. t 2 + t +
6.
4x 2 + 12x + 9
7.
16b 2 - 8b + 1
8.
9a 2 + 12a + 4
4x
4
+
3
9
6y
1
18. 9y 2 +
+
5
25
9.
25x 2 - 40x + 16
19. x 2 + 2 +
10. 49y 2 + 14y + 1
11. 9y 2 - 30y + 25
1
4
17. x 2 -
1
x2
20. 25k 2 - 20 +
4
k2
Chapter 2 Algebra and Surds
Difference of 2 squares
A special case of binomial products is ] a + b g ] a - b g = a 2 - b 2.
a2 - b2 = ] a + b g ] a - b g
EXAMPLES
Factorise
1. d 2 - 36
Solution
d 2 - 36 = d 2 - 6 2
= ]d + 6 g]d - 6 g
2. 9b 2 - 1
Solution
9b 2 - 1 = ] 3b g2 - 1 2
= ( 3 b + 1) ( 3 b - 1 )
3. (a + 3) 2 - (b - 1) 2
Solution
] a + 3 g2 - ] b - 1 g2 = [(a + 3) + (b - 1)] [(a + 3) - (b - 1)]
= (a + 3 + b - 1) ( a + 3 - b + 1)
= ( a + b + 2 ) (a - b + 4 )
2.12
Exercises
Factorise
1.
a2 - 4
7.
1 - 4z 2
2.
x2 - 9
8.
25t 2 - 1
3.
y2 - 1
9.
9t 2 - 4
4.
x 2 - 25
10. 9 - 16x 2
5.
4x 2 - 49
11. x 2 - 4y 2
6.
16y 2 - 9
12. 36x 2 - y 2
65
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Maths In Focus Mathematics Extension 1 Preliminary Course
13. 4a 2 - 9b 2
20.
14. x 2 - 100y 2
15. 4a - 81b
2
21. ] x + 2 g2 - ^ 2y + 1 h2
2
22. x 4 - 1
16. ]x + 2g2 - y 2
17. ] a - 1 g - ] b - 2 g
2
2
18. z - ] 1 + w g
2
19. x 2 -
y2
-1
9
2
1
4
23. 9x 6 - 4y 2
24. x 4 - 16y 4
25. a 8 - 1
Sums and differences of 2 cubes
a 3 + b 3 = ] a + b g ^ a 2 - ab + b 2 h
Proof
(a + b) (a 2 - ab + b 2) = a 3 - a 2 b + ab 2 + a 2 b - ab 2 + b 3
= a3 + b3
a 3 - b 3 = ] a - b g ^ a 2 + ab + b 2 h
Proof
(a - b) (a 2 + ab + b 2) = a 3 + a 2 b + ab 2 - a 2 b - ab 2 - b 3
= a3 - b3
EXAMPLES
Factorise
1. 8x 3 + 1
Solution
8x 3 + 1 = ] 2x g3 + 1 3
= (2x + 1) [] 2x g2 - (2x) (1) + 1 2]
= (2x + 1 ) (4 x 2 - 2 x + 1 )
Chapter 2 Algebra and Surds
2. 27a 3 - 64b 3
Solution
27a 3 - 64b 3 = ] 3a g3 - ] 4b g3
= (3a - 4b) [] 3a g2 + (3a) (4b) + ] 4b g2]
= (3a - 4b) (9a 2 + 12ab + 16b 2)
2.13
Exercises
Factorise
1.
b3 - 8
2.
x 3 + 27
3.
12.
x3
- 27
8
t3 + 1
13.
1000
1
+ 3
3
a
b
4.
a 3 - 64
14. ] x + 1 g3 - y 3
5.
1 - x3
15. 125x 3 y 3 + 216z 3
6.
8 + 27y 3
16. ]a - 2g3 - ]a + 1g3
7.
y 3 + 8z 3
8.
x 3 - 125y 3
9.
8x 3 + 27y 3
10. a 3 b 3 - 1
11. 1000 + 8t 3
17. 1 -
x3
27
18. y 3 + ]3 + xg3
19. ] x + 1 g3 + ^ y - 2 h3
20. 8]a + 3g3 - b 3
Mixed factors
Sometimes more than one method of factorising is needed to completely
factorise an expression.
EXAMPLE
Factorise 5x 2 - 45.
Solution
5x 2 - 45 = 5 (x 2 - 9)
= 5 (x + 3) (x - 3)
(using simple factors)
(the difference of two squares)
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Maths In Focus Mathematics Extension 1 Preliminary Course
2.14
Exercises
Factorise
1.
2x 2 - 18
16. x 3 - 3x 2 - 10x
2.
3p 2 - 3p - 36
17. x 3 - 3x 2 - 9x + 27
3.
5y 3 - 5
18. 4x 2 y 3 - y
4.
4a 3 b + 8a 2 b 2 - 4ab 2 - 2a 2 b
19. 24 - 3b 3
5.
5a 2 - 10a + 5
20. 18x 2 + 33x - 30
6.
- 2x 2 + 11x - 12
21. 3x 2 - 6x + 3
7.
3z 3 + 27z 2 + 60z
22. x 3 + 2x 2 - 25x - 50
8.
9ab - 4a 3 b 3
23. z 3 + 6z 2 + 9z
9.
x3 - x
24. 4x 4 - 13x 2 + 9
10. 6x 2 + 8x - 8
25. 2x 5 + 2x 2 y 3 - 8x 3 - 8y 3
11. 3m - 15 - 5n + mn
26. 4a 3 - 36a
12. ] x - 3 g2 - ] x + 4 g2
27. 40x - 5x 4
13. y 2 ^ y + 5 h - 16 ^ y + 5 h
28. a 4 - 13a 2 + 36
14. x 4 - x 3 + 8x - 8
29. 4k 3 + 40k 2 + 100k
15. x 6 - 1
30. 3x 3 + 9x 2 - 3x - 9
DID YOU KNOW?
Long division can be used to find factors of an expression. For example, x - 1 is a factor of
x 3 + 4x - 5. We can find the other factor by dividing x 3 + 4x - 5 by x - 1.
x2 + x + 5
x - 1 x3
+ 4x - 5
g
x3
-
x2
x 2 + 4x
x2
You will study this in
Chapter 12.
-
x
5x - 5
5x - 5
0
So the other factor of x 3 + 4x - 5 is x 2 + x + 5
` x 3 + 4x - 5 = (x - 1) (x 2 + x + 5)
Chapter 2 Algebra and Surds
69
Completing the Square
Factorising a perfect square uses the results
a 2 ! 2ab + b 2 = ] a ! b g2
EXAMPLES
1. Complete the square on x 2 + 6x.
Solution
Using a 2 + 2ab + b 2:
a=x
2ab = 6x
Substituting a = x:
2xb = 6x
b=3
Notice that 3 is half of 6.
To complete the square:
a 2 + 2ab + b 2 = ] a + b g2
2
x + 2x ] 3 g + 3 2 = ] x + 3 g2
x 2 + 6x + 9 = ] x + 3 g2
2. Complete the square on n 2 - 10n.
Solution
Using a 2 - 2ab + b 2:
a=n
2ab = 10x
Substituting a = n:
2nb = 10n
b=5
Notice that 5 is half of 10.
To complete the square:
a 2 - 2ab + b 2 = ] a - b g2
n 2 - 2n ] 5 g + 5 2 = ] n - 5 g2
n 2 - 10n + 25 = ] n - 5 g2
To complete the square on a 2 + pa, divide p by 2 and square it.
p 2
p 2
a 2 + pa + d n = d a + n
2
2
70
Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
1. Complete the square on x 2 + 12x.
Solution
Divide 12 by 2 and square it:
x 2 + 12x + c
12 2
m = x 2 + 12x + 6 2
2
= x 2 + 12x + 36
= ]x + 6g2
2. Complete the square on y 2 - 2y.
Solution
Divide 2 by 2 and square it:
2 2
y 2 - 2y + c m = y 2 - 2 y + 1 2
2
= y 2 - 2y + 1
= ^ y - 1 h2
2.15
Exercises
Complete the square on
1.
x 2 + 4x
12. y 2 + 3y
2.
b 2 - 6b
13. x 2 - 7x
3.
x 2 - 10x
14. a 2 + a
4.
y 2 + 8y
15. x 2 + 9x
5.
m 2 - 14m
16. y 2 -
6.
q 2 + 18q
5y
2
7.
x 2 + 2x
17. k 2 -
11k
2
8.
t 2 - 16t
18. x 2 + 6xy
9.
x 2 - 20x
19. a 2 - 4ab
10. w 2 + 44w
11. x 2 - 32x
20. p 2 - 8pq
Chapter 2 Algebra and Surds
71
Algebraic Fractions
Simplifying fractions
EXAMPLES
Simplify
4x + 2
2
1.
Solution
2 ] 2x + 1 g
4x + 2
=
2
2
= 2x + 1
Factorise first, then cancel.
2x 2 - 3x - 2
x3 - 8
2.
Solution
] 2x + 1 g ] x - 2 g
2x 2 - 3x - 2
=
3
] x - 2 g ^ x 2 + 2x + 4 h
x -8
2x + 1
= 2
x + 2x + 4
2.16
Exercises
Simplify
1.
5a + 10
5
2.
3.
4.
9.
b3 - 1
b2 - 1
6t - 3
3
10.
8y + 2
6
2p 2 + 7p - 15
6p - 9
11.
a2 - 1
a + 2a - 3
8
4d - 2
2
5.
6.
x
5x 2 - 2x
y-4
12.
13.
y - 8y + 16
2
3 ]x - 2g + y ]x - 2g
x3 - 8
x 3 + 3x 2 - 9x - 27
x 2 + 6x + 9
2
7.
2ab - 4a 2
a 2 - 3a
8.
s2 + s - 2
s 2 + 5s + 6
14.
15.
2p 2 - 3p - 2
8p 3 + 1
ay - ax + by - bx
2ay - by - 2ax + bx
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Maths In Focus Mathematics Extension 1 Preliminary Course
Operations with algebraic fractions
EXAMPLES
Simplify
1.
x+3
x-1
5
4
Solution
Do algebraic fractions
the same way as ordinary
fractions.
4 ]x - 1 g - 5 ]x + 3 g
x -1 x +3
=
5
4
20
4x - 4 - 5x - 15
=
20
- x - 19
=
20
2.
2a 2 b + 10ab
a 2 - 25
'
3
4b + 12
b + 27
Solution
2a 2 b + 10ab
a 2 - 25
2a 2 b + 10ab 4b + 12
'
=
# 2
4b + 12
b 3 + 27
b 3 + 27
a - 25
2ab ] a + 5 g
4 ]b + 3 g
=
#
2
]
a + 5 g]a - 5 g
] b + 3 g ^ b - 3b + 9 h
8ab
=
] a - 5 g ^ b 2 - 3b + 9 h
3.
2
1
+
x-5 x+2
Solution
2 ]x + 2g + 1 ]x - 5g
2
1
+
=
x-5 x+2
]x - 5g]x + 2g
2x + 4 + x - 5
=
]x - 5g]x + 2g
3x - 1
=
]x - 5g]x + 2g
Chapter 2 Algebra and Surds
2.17
1.
2.
Exercises
Simplify
x 3x
(a) +
4
2
y + 1 2y
(b)
+
5
3
a+2 a
(c)
4
3
p-3 p+2
(d)
+
6
2
x-5 x-1
(e)
2
3
4.
Simplify
3
b 2 + 2b
#
(a)
b + 2 6a - 3
1
1
+
x+1 x-3
(g)
3
2
x
2
+
x -4
(h)
1
1
+
a 2 + 2a + 1 a + 1
(i)
5
2
1
+
y+2 y+3 y-1
(j)
2
7
x 2 - 16 x 2 - x - 12
2
Simplify
(a)
y2 - 9
3x 2
x 2 - 2x - 8
#
#
4y - 12 6x - 24
y 3 + 27
q3 + 1
(b) 2
#
q + 2q + 1 p + 2
(b)
2
a 2 - 5a
3a - 15 y - y - 2
'
#
5ay
y 2 - 4y + 4
y2 - 4
3ab 2
12ab - 6a
(c)
' 2
5xy
x y + 2xy 2
(c)
3
x 2 + 3x
2x + 8
+ 2
#
x-3
4x - 16
x -9
(d)
5b
b2
b
' 2
2b + 6
b
1
+
b +b-6
(e)
x 2 - 8x + 15
x 2 - 9 x 2 + 5x + 6
'
#
2
2x - 10
5x + 10x
10x 2
p2 - 4
(d)
ax - ay + bx - by
x2 - y2
#
x3 + y3
ab 2 + a 2 b
x 2 - 6x + 9
x 2 - 5x + 6
(e)
'
x 2 - 25
x 2 + 4x - 5
3.
(f)
5.
Simplify
2 3
(a) x + x
Simplify
(a)
1
2
4
+
x 2 - 7x + 10 x 2 - 2x - 15 x 2 + x - 6
1
2
x-1 x
(b)
3
5
2
+
2
2
x
x
x -4
(c) 1 +
3
a+b
(c)
3
2
+
p 2 + pq pq - q 2
(d) x -
x2
x+2
(d)
a
b
1
+
a + b a - b a2 - b2
(b)
(e) p - q +
1
p+q
2
x+y
y
x
(e) x - y + y - x - 2
y - x2
Substitution
Algebra is used in writing general formulae or rules. For example, the formula
A = lb is used to find the area of a rectangle with length l and breadth b. We
can substitute any values for l and b to find the area of different rectangles.
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
1. P = 2l + 2b is the formula for finding the perimeter of a rectangle
with length l and breadth b. Find P when l = 1.3 and b = 3.2.
Solution
P = 2 l + 2b
= 2 ] 1 . 3 g + 2 ] 3 .2 g
= 2 .6 + 6 . 4
=9
2. V = rr 2 h is the formula for finding the volume of a cylinder with
radius r and height h. Find V (correct to 1 decimal place) when r = 2.1
and h = 8.7.
Solution
V = rr 2 h
= r ] 2.1 g2 (8.7)
= 120.5 correct to 1 decimal place
9C
+ 32 is the formula for changing degrees Celsius ] °C g into
5
degrees Fahrenheit ] °F g find F when C = 25.
3. If F =
Solution
9C
+ 32
5
9 ] 25 g
=
+ 32
5
225
=
+ 32
5
225 + 160
=
5
385
=
5
= 77
This means that 25°C is the same as 77°F.
F=
Chapter 2 Algebra and Surds
2.18
1.
Exercises
Given a = 3.1 and b = - 2.3 find,
correct to 1 decimal place.
(a) ab
(b) 3b
(c) 5a 2
(d) ab 3
(e) ]a + bg2
(f)
a-b
(g) - b 2
2.
T = a + ] n - 1 g d is the formula
for finding the term of an
arithmetic series. Find T when
a = - 4, n = 18 and d = 3.
3.
Given y = mx + b, the equation
of a straight line, find y if
m = 3, x = - 2 and b = - 1.
4.
If h = 100t - 5t 2 is the height of
a particle at time t, find h when
t = 5.
5.
Given vertical velocity v = - gt,
find v when g = 9.8 and t = 20.
6.
If y = 2 x + 3 is the equation of
a function, find y when x = 1.3,
correct to 1 decimal place.
7.
S = 2r r ] r + h g is the formula for
the surface area of a cylinder.
Find S when r = 5 and h = 7,
correct to the nearest whole
number.
8.
A = rr 2 is the area of a circle with
radius r. Find A when r = 9.5,
correct to 3 significant figures.
9.
n-1
Given u n = ar
is the nth term
of a geometric series, find u n if
a = 5, r = - 2 and n = 4.
10. Given V = 1 lbh is the volume
3
formula for a rectangular
pyramid, find V if l = 4.7, b = 5.1
and h = 6.5.
11. The gradient of a straight line is
y2 - y1
given by m = x - x . Find m
2
1
if x 1 = 3, x 2 = -1, y 1 = - 2 and
y 2 = 5.
12. If A = 1 h ] a + b g gives the area
2
of a trapezium, find A when
h = 7, a = 2.5 and b = 3.9.
13. Find V if V = 4 rr 3 is the volume
3
formula for a sphere with radius r
and r = 7.6, to 1 decimal place.
14. The velocity of an object at a
certain time t is given by the
formula v = u + at. Find v when
u = 1 , a = 3 and t = 5 .
4
5
6
a
15. Given S =
, find S if a = 5
1-r
and r = 2 . S is the sum to infinity
3
of a geometric series.
16. c = a 2 + b 2 , according to
Pythagoras’ theorem. Find the
value of c if a = 6 and b = 8.
17. Given y = 16 - x 2 is the
equation of a semicircle, find the
exact value of y when x = 2.
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Maths In Focus Mathematics Extension 1 Preliminary Course
18. Find the value of E in the energy
equation E = mc 2 if m = 8.3 and
c = 1.7.
19. A = P c 1 +
20. If S =
a geometric series, find S if
a = 3, r = 2 and n = 5.
r n
m is the formula
100
for finding compound interest.
Find A when P = 200, r = 12 and
n = 5, correct to 2 decimal places.
a ^rn - 1h
is the sum of
r -1
21. Find the value of
a3 b2
if
c2
2
3
1 4
a = c 3 m , b = c 2 m and c = c m .
4
3
2
Surds
An irrational number is a number that cannot be written as a ratio or fraction
(rational). Surds are special types of irrational numbers, such as 2, 3 and 5 .
Some surds give rational values: for example, 9 = 3. Others, like 2 , do
not have an exact decimal value. If a question involving surds asks for an exact
answer, then leave it as a surd rather than giving a decimal approximation.
Simplifying surds
Class Investigations
1. Is there an exact decimal equivalent for 2 ?
2. Can you draw a line of length exactly 2 ?
3. Do these calculations give the same results?
(a) 9 # 4 and 9 # 4
(b)
4
and
4
9
(c)
9
9 + 4 and
9 +
4
(d)
9 - 4 and
9 -
4
Here are some basic properties of surds.
a# b =
ab
a' b =
a
^ x h2 =
b
=
x2 = x
a
b
Chapter 2 Algebra and Surds
77
EXAMPLES
1. Express in simplest surd form
45 .
45 also equals
3 # 15 but this will
not simplify. We look
for a number that is a
perfect square.
Solution
45 = 9 # 5
= 9 # 5
=3# 5
=3 5
2. Simplify 3 40 .
Solution
Find a factor of 40 that
is a perfect square.
3 40 = 3 4 # 10
= 3 # 4 # 10
= 3 # 2 # 10
= 6 10
3. Write 5 2 as a single surd.
Solution
5 2 =
=
2.19
1.
25 # 2
50
Exercises
Express these surds in simplest
surd form.
(k)
112
(l)
300
(a)
12
(b)
63
(c)
24
(d)
50
(e)
72
(f)
200
(g)
48
(h)
75
(i)
32
(a) 2 27
(j)
54
(b) 5 80
(m) 128
2.
(n)
243
(o)
245
(p)
108
(q)
99
(r)
125
Simplify
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Maths In Focus Mathematics Extension 1 Preliminary Course
(c) 4 98
(g) 3 13
(d) 2 28
(h) 7 2
(e) 8 20
(i) 11 3
(f) 4 56
(j) 12 7
(g) 8 405
4.
(h) 15 8
(a)
(i) 7 40
x =3 5
(b) 2 3 =
x
(c) 3 7 =
x
Write as a single surd.
(d) 5 2 =
x
(a) 3 2
(e) 2 11 =
(b) 2 5
(f)
(c) 4 11
(g) 4 19 =
(d) 8 2
(h)
(e) 5 3
(i) 5 31 =
(f) 4 10
(j)
(j) 8 45
3.
Evaluate x if
x
x =7 3
x
x = 6 23
x
x = 8 15
Addition and subtraction
Calculations with surds are similar to calculations in algebra. We can only add
or subtract ‘like terms’ with algebraic expressions. This is the same with surds.
EXAMPLES
1. Simplify 3 2 + 4 2 .
Solution
3 2+4 2 =7 2
2. Simplify
3 - 12 .
Solution
First, change into ‘like’ surds.
3 - 12 = 3 - 4 # 3
= 3 -2 3
=- 3
3. Simplify 2 2 - 2 + 3 .
Solution
2 2- 2+ 3=
2+ 3
Chapter 2 Algebra and Surds
2.20
79
Exercises
Simplify
1.
5 +2 5
14.
50 -
32
2.
3 2 -2 2
15.
28 +
63
3.
3 +5 3
16. 2 8 -
18
4.
7 3 -4 3
5.
5 -4 5
4 6 -
6.
17. 3 54 + 2 24
18.
90 - 5 40 - 2 10
19. 4 48 + 3 147 + 5 12
6
7.
2 -8 2
20. 3 2 + 8 - 12
8.
5 +4 5 +3 5
21.
63 - 28 - 50
9.
2 -2 2 -3 2
22.
12 - 45 - 48 - 5
10.
5 +
45
23.
150 + 45 + 24
11.
8 -
2
24.
32 - 243 - 50 + 147
12.
3 +
48
25.
80 - 3 245 + 2 50
13.
12 -
27
Multiplication and division
To get a b # c d = ac bd ,
multiply surds with surds and
rationals with rationals.
a # b = ab
a b # c d = ac bd
a# a =
a
b
=
a2 = a
a
b
EXAMPLES
Simplify
1. 2 2 #- 5 7
Solution
2 2 #- 5 7 = -10 14
CONTINUED
80
Maths In Focus Mathematics Extension 1 Preliminary Course
2. 4 2 # 5 18
Solution
4 2 # 5 18 = 20 36
= 20 # 6
= 120
3.
2 14
4 2
Solution
2 14
4 2
=
2 2 #
7
2
=
4.
7
4 2
3 10
15 2
Solution
3 10
15 2
=
3# 5 # 2
15 2
5
=
5
5. d
2
10 n
3
Solution
2
^ 10 h
10 n
=
3
^ 3 h2
10
=
3
=31
3
2
d
Chapter 2 Algebra and Surds
2.21
Exercises
Simplify
1.
7 #
2.
3# 5
3.
2 #3 3
4.
5 7 #2 2
3
5.
-3 3 #2 2
6.
5 3 #2 3
7.
- 4 5 # 3 11
8.
2 7# 7
9.
2 3 # 5 12
10.
6# 2
11.
8 #2 6
23.
24.
25.
26.
27.
28.
5 8
10 2
16 2
2 12
10 30
5 10
2 2
6 20
4 2
8 10
3
3 15
2
29.
8
12. 3 2 # 5 14
13.
10 # 2 2
14. 2 6 #-7 6
15. ^ 2 h
2
2
16. ^ 2 7 h
17.
31.
32.
3 15
6 10
5 12
5 8
15 18
10 10
3# 5# 2
18. 2 3 # 7 #- 5
19.
30.
2 # 6 #3 3
33.
15
2 6
2n
3
35. d
5n
7
20. 2 5 # - 3 2 # - 5 5
21.
22.
4 12
2 2
2
34. d
2
12 18
3 6
Expanding brackets
The same rules for expanding brackets and binomial products that you use in
algebra also apply to surds.
81
82
Maths In Focus Mathematics Extension 1 Preliminary Course
Simplifying surds by removing grouping symbols uses these general rules.
a^ b + ch=
ab + ac
Proof
a^ b + ch =
=
a# b +
ab + ac
a# c
Binomial product:
^ a + b h^ c + d h =
ac +
ad +
bc +
bd
Proof
^ a + b h^ c + d h = a # c + a # d + b # c + b # d
= ac + ad + bc + bd
Perfect squares:
^ a + b h2 = a + 2 ab + b
Proof
^ a + b h2 = ^ a + b h ^ a + b h
= a 2 + ab + ab + b 2
= a + 2 ab + b
^ a - b h2 = a - 2 ab + b
Proof
^ a - b h2 = ^ a - b h ^ a - b h
= a 2 - ab - ab + b 2
= a - 2 ab + b
Difference of two squares:
^ a + b h^ a - b h = a - b
Proof
^ a + b h ^ a - b h = a 2 - ab + ab - b 2
=a-b
Chapter 2 Algebra and Surds
83
EXAMPLES
Expand and simplify
1. 2 ^ 5 + 2 h
Solution
2( 5 +
2) =
=
=
2# 5 +
10 + 4
10 + 2
2# 2
2. 3 7 ^ 2 3 - 3 2 h
Solution
3 7 (2 3 - 3 2 ) = 3 7 # 2 3 - 3 7 # 3 2
= 6 21 - 9 14
3. ^ 2 + 3 5 h ^ 3 -
2h
Solution
( 2 + 3 5)( 3 -
2) =
=
2# 3 - 2# 2 +3 5# 3 -3 5# 2
6 - 2 + 3 15 - 3 10
4. ^ 5 + 2 3 h ^ 5 - 2 3 h
Solution
( 5 + 2 3 ) ( 5 - 2 3 ) = 5 # 5 - 5 #2 3 + 2 3 # 5 - 2 3 #2 3
= 5 - 2 15 + 2 15 - 4#3
= 5 - 12
= -7
Another way to do this question is by using the difference of two squares.
2
2
( 5 + 2 3)( 5 - 2 3) = ^ 5 h - ^2 3 h
= 5 - 4#3
= -7
Notice that using the
difference of two
squares gives a rational
answer.
84
Maths In Focus Mathematics Extension 1 Preliminary Course
2.22
1.
Exercises
(m)^ 2 11 + 5 2 h^ 2 11 - 5 2 h
Expand and simplify
(a)
2^ 5 + 3h
(b)
3 ^2 2 - 5 h
(n) ^ 5 + 2 h
2
2
(o) ^ 2 2 - 3 h
(c) 4 3 ^ 3 + 2 5 h
(d)
2
(p) ^ 3 2 + 7 h
7 ^5 2 - 2 3 h
2
(q) ^ 2 3 + 3 5 h
(e) - 3 ^ 2 - 4 6 h
(f)
2
(r) ^ 7 - 2 5 h
3 ^ 5 11 + 3 7 h
2
(s) ^ 2 8 - 3 5 h
(g) - 3 2 ^ 2 + 4 3 h
(h)
5^ 5 - 5 3h
(i)
3 ^ 12 + 10 h
2
(t) ^ 3 5 + 2 2 h
3.
If a = 3 2 , simplify
(a) a2
(b) 2a3
(c) (2a)3
(d) ]a + 1g2
(e) ] a + 3 g ] a – 3 g
4.
Evaluate a and b if
2
(a) ^ 2 5 + 1h = a + b
(j) 2 3 ^ 18 + 3 h
(k) - 4 2 ^ 2 - 3 6 h
(l) - 7 5 ^ - 3 20 + 2 3 h
(m) 10 3 ^ 2 - 2 12 h
(n) - 2 ^ 5 + 2 h
(o) 2 3 ^ 2 - 12 h
2.
(b) ^ 2 2 - 5 h ^ 2 - 3 5 h
= a + b 10
Expand and simplify
(a) ^ 2 + 3h^ 5 + 3 3 h
5.
Expand and simplify
(a) ^ a + 3 - 2 h ^ a + 3 + 2 h
2
(b) _ p - 1 - p i
6.
Evaluate k if
^ 2 7 - 3 h ^ 2 7 + 3 h = k.
(g) ^ 7 + 3 h^ 7 - 3 h
7.
Simplify _ 2 x + y i _ x - 3 y i .
(h) ^ 2 - 3 h^ 2 + 3 h
8.
If ^ 2 3 - 5 h = a - b , evaluate
a and b.
9.
Evaluate a and b if
^ 7 2 - 3 h2 = a + b 2 .
(b) ^ 5 - 2 h^ 2 - 7 h
(c) ^ 2 + 5 3 h^ 2 5 - 3 2 h
(d) ^ 3 10 - 2 5 h^ 4 2 + 6 6 h
(e) ^ 2 5 - 7 2 h^ 5 - 3 2 h
(f) ^ 5 + 6 2 h^ 3 5 - 3 h
(i) ^ 6 + 3 2 h^ 6 - 3 2 h
(j) ^ 3 5 + 2 h^ 3 5 - 2 h
(k) ^ 8 - 5 h^ 8 + 5 h
(l) ^ 2 + 9 3 h^ 2 - 9 3 h
2
10. A rectangle has sides 5 + 1 and
2 5 - 1. Find its exact area.
Rationalising the denominator
Rationalising the denominator of a fractional surd means writing it with a
rational number (not a surd) in the denominator. For example, after
3 5
3
rationalising the denominator,
becomes
.
5
5
Chapter 2 Algebra and Surds
85
DID YOU KNOW?
A major reason for rationalising the denominator used to be to make it easier to evaluate the
fraction (before calculators were available). It is easier to divide by a rational number than an
irrational one; for example,
3
= 3 ' 2.236
5
3
5
5
This is hard to do
without a calculator.
This is easier to calculate.
= 3 # 2.236 ' 5
Squaring a surd in the denominator will rationalise it since ^ x h = x.
2
Multiplying by
b
a b
a
#
=
b
b
b
b
b
is the same as
multiplying by 1.
Proof
b
a b
a
#
=
b
b
b2
a b
=
b
EXAMPLES
1. Rationalise the denominator of
3
.
5
Solution
5
3 5
3
#
=
5
5
5
2. Rationalise the denominator of
Solution
2
5 3
.
Don’t multiply by
5
2
5 3
#
3
3
=
2 3
5 9
2 3
=
5# 3
2 3
=
15
3
as it takes
5 3
longer to simplify.
86
Maths In Focus Mathematics Extension 1 Preliminary Course
When there is a binomial denominator, we use the difference of two
squares to rationalise it, as the result is always a rational number.
To rationalise the denominator of
a+ b
c+ d
, multiply by
Proof
a+ b
c+ d
^ a + b h^ c - d h
c- d
^ c + d h^ c - d h
^ a + b h^ c - d h
=
^ c h2 - ^ d h2
^ a + b h^ c - d h
=
c-d
c- d
#
=
EXAMPLES
1. Write with a rational denominator
5
2 -3
Multiply by the conjugate
surd 2 + 3.
.
Solution
5
2 -3
2 +3
#
2 +3
=
5 ^ 2 + 3h
^ 2 h2 - 3 2
10 + 3 5
=
2-9
10 + 3 5
=
-7
10 + 3 5
=7
2. Write with a rational denominator
2 3+ 5
3+4 2
.
Solution
2 3 +
5
3 +4 2
#
3 -4 2
3 -4 2
=
^2 3 + 5 h^ 3 - 4 2 h
^ 3 h2 - ^ 4 2 h2
2 # 3 - 8 6 + 15 - 4 10
=
3 - 16 # 2
c- d
c- d
Chapter 2 Algebra and Surds
6 - 8 6 + 15 - 4 10
- 29
- 6 + 8 6 - 15 + 4 10
=
29
=
3. Evaluate a and b if
3 3
3- 2
= a + b.
Solution
3 3
3- 2
#
3+ 2
3+ 2
=
3 3^ 3 + 2h
^ 3 - 2 h^ 3 + 2 h
3 9+3 6
=
^ 3 h2 - ^ 2 h2
3#3+3 6
3-2
9+3 6
=
1
=9+3 6
=
=9+ 9# 6
= 9 + 54
So a = 9 and b = 54.
4. Evaluate as a fraction with rational denominator
2
+
3+2
5
3-2
.
Solution
2
+
3+2
5
3 -2
=
2^ 3 - 2h + 5 ^ 3 + 2h
^ 3 + 2h ^ 3 - 2h
2 3 - 4 + 15 + 2 5
=
^ 3 h2 - 2 2
2 3 - 4 + 15 + 2 5
3-4
2 3 - 4 + 15 + 2 5
=
-1
= - 2 3 + 4 - 15 - 2 5
=
87
88
Maths In Focus Mathematics Extension 1 Preliminary Course
2.23
1.
Express with rational
denominator
(a)
(b)
(c)
(d)
(e)
2.
Exercises
3.
1
7
(a)
3
(b)
2 2
2 3
(c)
5
6 7
(d)
5 2
1+
2
3
6 -5
(g)
5 +2 2
8+3 2
(j)
4 3 -2 2
(f)
1
5 +
2
2 -
7
2 +
3
2 +3
4 5
(j)
7 5
(k)
4
3 +
(l)
2
3
4.
2 -7
5 +2 6
3 -4
3 +4
3 3 3 +
(b)
(c)
2 +5
2
2
2 5 +3 2
3
2 +
+
3
3 2 -
#
3
6 -
3
2
3
2 +3
5
6 +2
2 +7
4+
2 3 +
3 -2
3
6 +
1
3
+
2
-
2
2
3 -
2
(d)
(e)
2 5
3
4 2
2
2 -1
+
+
5 -
3
5
2
3
3
5 3
2
4-
3
2+
3
3 +1
Find a and b if
(a)
2 3
-
1
where z = 1 +
z2
(h)
(i)
1
2 -1
1
where t =
t
3 2 +4
2 7
Express with rational
denominator
(e)
3
(g)
5
(i)
(d)
2 -
2
3 2 -4
(c)
2
(f) z 2 -
(h)
(b)
1
+
2 +1
(e) t +
(f)
(a)
Express as a single fraction with
rational denominator
=
a
b
=
a 6
b
2
=a+b 5
5 +1
2 7
7 -4
2 +3
2 -1
=a+b 7
=a+
b
2
-
2
6 -1
Chapter 2 Algebra and Surds
5.
2 -1
Show that
2 +1
+
4
is
2
7.
If x =
(b) x 2 +
2
+
1
5 -
2
-
as a single fraction with
3
rational denominator.
3 + 2, simplify
1
(a) x + x
2
5 +
5 +1
rational.
6.
Write
8.
1
x2
Show that
8
2
+
is
3+2 2
2
rational.
2
1
(c) b x + x l
9.
1
If 2 + x = 3 , where x ! 0,
find x as a surd with rational
denominator.
10. Rationalise the denominator of
b +2
]b ! 4 g
b -2
89
90
Maths In Focus Mathematics Extension 1 Preliminary Course
Test Yourself 2
1.
2.
3.
4.
Simplify
(a) 5y - 7y
3a + 12
(b)
3
(c) - 2k 3 # 3k 2
y
x
(d) +
5
3
(e) 4a - 3b - a - 5b
(f) 8 + 32
(g) 3 5 - 20 + 45
Factorise
(a) x 2 - 36
(b) a 2 + 2a - 3
(c) 4ab 2 - 8ab
(d) 5y - 15 + xy - 3x
(e) 4n - 2p + 6
(f) 8 - x 3
Expand and simplify
(a) b + 3 ] b - 2 g
(b) ] 2x - 1 g ] x + 3 g
(c) 5 ] m + 3 g - ] m - 2 g
(d) ]4x - 3g2
(e) ^ p - 5h^ p + 5h
(f) 7 - 2 ] a + 4 g - 5a
(g) 3 ^ 2 2 - 5 h
(h) ^ 3 + 7 h^ 3 - 2h
Simplify
4a - 12
10b
(a)
# 3
5b 3
a - 27
(b)
5.
5m + 10
m2 - 4
'
2
m - m - 2 3m + 3
The volume of a cube is V = s 3.
Evaluate V when s = 5.4.
6.
(a) Expand and simplify
^ 2 5 + 3 h ^ 2 5 - 3 h.
(b) Rationalise the denominator of
3 3
.
2 5+ 3
7.
Simplify
8.
If a = 4, b = - 3 and c = - 2, find the
value of
(a) ab 2
(b) a - bc
(c) a
(d) ]bcg3
(e) c ] 2a + 3b g
9.
Simplify
3 12
(a)
6 15
(b)
3
1
2
+
- 2
.
x-2
x+3
x +x-6
4 32
2 2
10. The formula for the distance an object
falls is given by d = 5t 2 . Find d when
t = 1.5.
11. Rationalise the denominator of
2
(a)
5 3
(b)
1+ 3
2
12. Expand and simplify
(a) ^ 3 2 - 4h^ 3 - 2 h
2
(b) ^ 7 + 2h
13. Factorise fully
(a) 3x 2 - 27
(b) 6x 2 - 12x - 18
(c) 5y 3 + 40
Chapter 2 Algebra and Surds
14. Simplify
3x 4 y
(a)
9xy 5
(b)
5
15x - 5
15. Simplify
2
(a) ^ 3 11 h
3
(b) ^ 2 3 h
16. Expand and simplify
(a) ] a + b g ] a - b g
(b) ] a + b g 2
(c) ] a - b g 2
17. Factorise
(a) a 2 - 2ab + b 2
(b) a 3 - b 3
1
18. If x = 3 + 1, simplify x + x and
give your answer with a rational
denominator.
19. Simplify
4 3
(a) a +
b
(b)
x-3 x-2
5
2
20. Simplify
2
3
, writing
5+2 2 2-1
your answer with a rational denominator.
21. Simplify
(a) 3 8
(b) - 2 2 # 4 3
(c) 108 - 48
(d)
23. Rationalise the denominator of
3
(a)
7
(b)
2
5 3
2
(c)
5 -1
(d)
(e)
2 2
3 2+ 3
5+ 2
4 5-3 3
24. Simplify
3x
x-2
(a)
5
2
a+2
2a - 3
(b)
+
7
3
1
2
(c) 2
1
x
+
x -1
4
1
(d) 2
+
k + 2k - 3 k + 3
(e)
3
2+ 5
-
5
3- 2
25. Evaluate n if
(a) 108 - 12 =
(b)
112 + 7 =
n
n
8 6
(c) 2 8 + 200 =
2 18
(d) 4 147 + 3 75 = n
180
(e) 2 245 +
= n
2
(e) 5a # - 3b # - 2a
(f)
22. Expand and simplify
(a) 2 2 ^ 3 + 2 h
(b) ^ 5 7 - 3 5 h^ 2 2 - 3 h
(c) ^ 3 + 2 h^ 3 - 2 h
(d) ^ 4 3 - 5 h^ 4 3 + 5 h
2
(e) ^ 3 7 - 2 h
2m 3 n
6m 2 n 5
(g) 3x - 2y - x - y
n
91
92
Maths In Focus Mathematics Extension 1 Preliminary Course
26. Evaluate x 2 +
1+2 3
1
if x =
2
x
1-2 3
27. Rationalise the denominator of
3
2 7
(there may be more than one answer).
21
(a)
28
2 21
(b)
28
21
(c)
14
21
(d)
7
x-3
x +1
.
5
4
-]x + 7 g
20
x+7
20
x + 17
20
- ] x + 17 g
20
28. Simplify
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
32. Simplify 5ab - 2a 2 - 7ab - 3a 2 .
(a) 2ab + a 2
(b) - 2ab - 5a 2
(c) - 13a 3 b
(d) - 2ab + 5a 2
33. Simplify
(a)
(b)
(c)
29. Factorise x 3 - 4x 2 - x + 4 (there may be
more than one answer).
(a) ^ x 2 - 1 h ] x - 4 g
(b) ^ x 2 + 1 h ] x - 4 g
(c) x 2 ] x - 4 g
(d) ] x - 4 g ] x + 1 g ] x - 1 g
30. Simplify 3 2 + 2 98 .
(a) 5 2
(b) 5 10
(c) 17 2
(d) 10 2
3
2
1
+
.
x-2
x+2
x2 - 4
x+5
]x + 2g]x - 2g
x+1
]x + 2g]x - 2g
x+9
]x + 2g]x - 2g
x-3
]x + 2g]x - 2g
31. Simplify
(d)
80
.
27
4 5
3 3
4 5
9 3
8 5
9 3
8 5
3 3
34. Expand and simplify ^ 3x - 2y h2 .
(a) 3x 2 - 12xy - 2y 2
(b) 9x 2 - 12xy - 4y 2
(c) 3x 2 - 6xy + 2y 2
(d) 9x 2 - 12xy + 4y 2
35. Complete the square on a 2 - 16a.
(a) a 2 - 16a + 16 = ^ a - 4 h2
(b) a 2 - 16a + 64 = ^ a - 8 h2
(c) a 2 - 16a + 8 = ^ a - 4 h2
(d) a 2 - 16a + 4 = ^ a - 2 h2
Chapter 2 Algebra and Surds
Challenge Exercise 2
1.
2.
Expand and simplify
(a) 4ab ] a - 2b g - 2a 2 ] b - 3a g
(b) _ y 2 - 2 i_ y 2 + 2 i
(c) ] 2x - 5 g3
Find the value of x + y with rational
denominator if x = 3 + 1 and
1
y=
.
2 5-3
2 3
2x + y
x-y
3x + 2y
.
+
- 2
x-3
x+3
x +x-6
12. (a) Expand ^ 2x - 1 h3.
6x 2 + 5x - 4
(b) Simplify
.
8x 3 - 12x 2 + 6x - 1
13. Expand and simplify ] x - 1 g ^ x - 3 h2.
14. Simplify and express with rational
2 +
5
-
5 3
3.
Simplify
4.
b
Complete the square on x 2 + a x.
15. Complete the square on x 2 + 2 x.
3
Factorise
(a) (x + 4)2 + 5 (x + 4)
(b) x 4 - x 2 y - 6y 2
(c) 125x 3 + 343
(d) a 2 b - 2a 2 - 4b + 8
16. If x =
5.
6.
7.
8.
9.
7 6 - 54
.
11. Simplify
denominator
Simplify
d=
4x 2 - 16x + 12
| ax 1 + by 1 + c |
.
Simplify
10. Factorise
^a + 1h
a3 + 1
.
a2
4
- 2.
2
x
b
.
lx 1 + kx 2
17. Find the exact value with rational
1
denominator of 2x 2 - 3x + x if x = 2 5 .
18. Find the exact value of
1+2 3
1
(a) x 2 + 2 if x =
x
1-2 3
(b) a and b if
is the formula for
a2 + b2
the perpendicular distance from a
point to a line. Find the exact value
of d with a rational denominator if
a = 2, b = -1, c = 3, x 1 = - 4 and y 1 = 5.
3
2 -1
, find the value of x when
k+l
k = 3, l = - 2, x 1 = 5 and x 2 = 4.
Complete the square on 4x 2 + 12x.
2xy + 2x - 6 - 6y
3 +4
3 -4
2+3 3
=a+b 3
19. A = 1 r 2 i is the area of a sector of a
2
circle. Find the value of i when A = 12
and r = 4.
20. If V = rr 2 h is the volume of a cylinder,
find the exact value of r when V = 9 and
h = 16.
21. If s = u + 1 at 2, find the exact value of s
2
when u = 2, a = 3 and t = 2 3 .
93