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SAMPLE
CHAPTERS
chapter
Real Numbers
Learning Outcomes
In this chapter you will learn:
 How to define the real numbers
 About factors, multiples and prime factors
 How to write a whole number as a product of prime factors
 The properties of integers
 The properties of rational numbers
 About rounding and significant figures
__
__
 How to geometrically construct √ 2 and √ 3
 About orders of magnitude and scientific notation
REAL NUMBERS
1
–8 –7 –6 –5 –4 –3 –2 –1
0
1
2
3
4
5
6
7
8
This is not a very rigorous definition of the real numbers. However, it will
serve our purposes. The discovery of a proper rigorous definition of the real
numbers was one of the most important developments in the mathematics
of the nineteenth century. The main contributors to the field were a French
mathematician, Augustin-Louis Cauchy (1789–1857), and two German
mathematicians, Richard Dedekind (1831–1916) and Karl Weierstrass
(1815–1897).
Augustin-Louis
Cauchy (1789–1857)
Richard Dedekind
(1831–1916)
Karl Weierstrass
(1815–1897)
The natural numbers, the integers, the rational numbers and
the irrational numbers are all subsets of the real number system.
1.1 FACTORS, MULTIPLES
AND PRIME
PRIM E FACTORS
Natural Numbers
Rationals
The natural numbers are the ordinary counting numbers.
The set of natural numbers is an infinite set. This means that
the set is never-ending. The letter N is used to label the set
of natural numbers.
Reals
Irrationals
Integers
Naturals
Factors
For example, all the factors of 24 are {1, 2, 3, 4, 6, 8, 12, 24}
As you can see, 1 is a factor of 24 and 24 is a factor of 24.
2
1
Multiples
REAL NUMBERS
The multiples of 6 are {6, 12, 18, 24, 30, 36 ...}
As you can see, the set of multiples is an infinite set, i.e. it goes on forever.
Prime Numbers
„ 7 is a prime number as it has two factors only: 1 and 7.
„ 2 is the only even prime number. Its two factors are 1 and 2.
„ 11 is the first two-digit prime. Its two factors are 1 and 11.
„ 1 is not a prime as it has one factor only, itself.
„ 0 is not a prime as it has an infinite number of factors, and it is not a natural number.
Euclid
The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic is an important result that shows that the primes are the
building blocks of the natural numbers.
3
1
REAL NUMBERS
Highest Common Factor (HCF) and Lowest Common
Multiple (LCM)
The highest common factor of 12 and 20 is 4, as 4 is the largest natural number that divides evenly
into both 12 and 20.
The lowest common multiple of 3 and 4 is 12, as 12 is the
smallest number that both 3 and 4 divide evenly into.
In Activities 1.1 and 1.2 you discovered how to find the HCF and LCM of two natural numbers,
by writing each number as a product of primes.
R
Wo rke d Exa m p le 1 . 1
Express 240 as a product of prime numbers.
Solution
2 240
2 120
Start with the lowest prime
that is a factor.
R
Worke d E x a m p l e 1. 2
Find (i) the HCF and (ii) the LCM of 512
and 280.
Solution
Express both numbers as a product of primes.
2 60
2 512
2 280
2 30
2 256
2 140
3 15
2 128
2 70
5 5
2 64
5 35
1
2 32
7 7
240 = 2 × 2 × 2 × 2 × 3 × 5
2 16
1
∴ 240 = 24 × 3 × 5
2 8
2 4
2 2
1
512 = 29 and 280 = 23 × 5 × 7
(i) HCF (512, 280) = 23 = 8
(ii) LCM (512, 280) = 29 × 5 × 7 = 17,920
4
R
1
Worke d Exa m p le 1 . 3
REAL NUMBERS
Show that, if p is a prime number and p divides evenly into r 2, then p divides evenly into r.
Solution
Let r = r 1a 1r 2a 2r 3a 3 ... r na n, where each rî is prime and each aî ∈ N.
Therefore, using the rules of indices:
r2 = (r 1α 1r 2α 2r 3α 3 ... r nα n)2
= r 12a 1r 22a 2r 32a 3 ... r n2a n
If p | r2 (p divides r2), then p | one of rî2aî.
Since each rî is prime, ⇒ p | one of rî.
Hence, p divides r 1a 1r 2a 2r 3a 3 ... r na n = r.
R
Worke d Exa m p le 1 . 4
Periodical cicadas are insects with very long larval periods and brief adult lives.
For each species of periodical cicada with larval period of 17 years, there is a
similar species with larval period of 13 years.
If both the 17-year and 13-year species emerged in a particular location in 2011,
when will they next both emerge in that location?
Solution
Here we are looking for the LCM of 17 and 13. As both numbers are prime the LCM of 13 and 17 is
17 × 13 = 221.
Therefore, both species will next appear together in that location in the year 2232.
R
Worke d Exa m p le 1 . 5
Evaluate (i) 5! and (ii) 5 × 4!
Solution
(i)
5! = 5 × 4 × 3 × 2 × 1
∴ 5! = 120
(ii) 5 × 4! = 5 × (4 × 3 × 2 × 1)
= 120
5
1
REAL NUMBERS
Infinitude of Primes
There are infinitely many prime numbers.
In the next activity you will discover a proof,
by contradiction, that this statement is correct.
Exercise 1.1
1. Express the following numbers as a product of
prime factors:
(i) 160
(v) 1,155
(ii) 273
(vi) 1,870
(iii) 128
(vii) 10,500
(viii) 102
(ix) 1,224
(x) 38,016
2. (a) Express each of the following pairs of
numbers as the product of prime factors.
(b) Hence, find the LCM and HCF for each set
of numbers.
(i) 102 and 170
(vi) 123 and 615
(ii) 117 and 130
(vii) 69 and 123
(iii) 368 and 621 (viii) 20, 30 and 60
(v) 60 and 765
(ix) 8, 10 and 20
(x) 294, 252 and 210
3. Let n be a natural number.
What is the HCF of:
(i) n and 2n
2
(ii) n and n ?
4. If 3 divides 3346112, explain why 3 must also
divide 334,611.
5. r is an even natural number and s is an odd
natural number. The prime factorisations of r
and s are:
r = r 1α 1r 2α 2r 3α 3... r nα n
s = s 1θ 1s 2θ 2s 3θ 3... s mθ m
(i) Explain why one of the primes,
r 1, r 2, r 3, ... r n must be 2.
(ii) Explain why none of the primes,
s 1, s 2, s 3, ... s m is 2.
6
u = u 1φ 1u 2φ 2u 3φ 3... u nφ n
(i) Explain why none of the primes,
u 1, u 2, u 3, ... u n divide u + 1.
(ii) Hence, write down the HCF of u and
u + 1.
(iv) 368
(iv) 58 and 174
6. Let u be any natural number with prime
factorisation,
7. Let u be any natural number with prime
factorisation,
u = u 1φ 1u 2φ 2u 3φ 3... u nφ n
(i) If u is even, then find the HCF of u and
u + 2.
(ii) If u is odd, then what is the HCF of u and
u + 2.
8. Kate has two pieces of material. One piece
is 72 cm wide and the other piece is 90 cm
wide. She wants to cut both pieces into strips
of equal width that are as wide as possible.
How wide should she cut the strips?
9. Tom exercises every 14 days and Katie every
nine days. Tom and Katie both exercised
on March 12. On what date will they both
exercise together again?
10. Miss Hoover has 160 crayons and 30 colouring
books to give to her students. If each student
gets an equal number of crayons and an equal
number of colouring books what is the largest
number of students she can have in her class?
11. Bart is making a board game with dimensions of
16 cm by 25 cm. He wants to use square tiles.
What are the dimensions of the largest tile he
can use?
(ii) Explain why F(n) is always a natural
number.
(iii) Consider the sequence:
n, F(n), F(F(n), F(F(F(n))) ....
Construct the first 15 terms of the
sequence for each of the following values
of n:
13. The Ulam numbers, u n, n ∈ N, are defined as
follows:
u 1 = 1,
u2 = 2
and each successive natural number, m,
where m > 2, u m is an Ulam number if and
only if it can be written uniquely as the sum
of two distinct Ulam numbers.
6, 10, 15, 32 and 17.
(iv) Based on your results make a conjecture
about the sequence.
(v) Test your conjecture for n = 39.
15. Consider the finite sequence:
10! + 2, 10! + 3, 10! + 4, ... 10! + 9, 10! + 10
(i) Explain the words in bold.
(i) How many terms are in the sequence?
(ii) Why is 5 not an Ulam number?
(ii) Explain why none of the terms are prime.
(iii) Find the first 10 Ulam numbers.
(iii) Construct a sequence of 20 consecutive
natural numbers none of which are
prime.
14. Let n be a natural number. Define the
function, F(n) as follows,
n
if n is even
F(n) = __
2
3n + 1
F(n) = _______ if n is odd
2
1
(i) Find F(1) and F(2).
REAL NUMBERS
12. Beginning on Monday of each week and
running until Friday, The Breakfast Show gives
away €100 to every 100th caller who gets
through to the show. During the week before
a Saturday night concert, the show offers two
free tickets to the concert for every 70th caller.
How many callers must get through before
one wins the tickets and the €100?
(iv) Is it possible to have a set of consecutive
natural numbers of any given size, that
does not contain any prime numbers?
Explain.
1.2 INTEGERS AND R ATIONAL NUM
NUMBERS
BERS
The integers are made up of zero and all the positive and negative whole numbers. Mathematicians use
the letter Z to represent the set of integers.
Properties of Integers
„ a + b and a × b are integers whenever a and b are integers. (Closure Property)
„ a + b = b + a and a × b = b × a (Commutative Properties)
„ (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c) (Associative Properties)
„ a × (b + c) = (a × b) + (a × c) (Distributive Property)
„ a + 0 = a and a × 1 = a for all integers a. (Identity Elements)
„ For every integer a, there exists an integer –a, such that a + (–a) = 0.
We say –a is the additive inverse of a. (Additive Inverse)
7
1
Rational Numbers
REAL NUMBERS
The letter Q is used to represent the set of rational numbers.
„ Rational numbers are also called fractions.
„ All integers can be written in the form _ba
and hence are rational numbers.
For example, 3 = _31.
„ The properties listed for the integers also carry over into the rationals.
„ Two fractions are equivalent if they have the same value. For example, _12 = _36.
„ The reciprocal of a fraction is found by turning the fraction upside down (inverting).
11
12
For example, the reciprocal of __
is __
.
12
11
„ Every fraction, with the exception of zero, has a multiplicative inverse. The product of a number
and its multiplicative inverse is always 1. The multiplicative inverse of a fraction is its reciprocal.
3 4
For example, the multiplicative inverse of __ is __.
4 3
R
Wo rke d Exa m p le 1 . 6
Use the properties of the integers to prove that for any integer a × 0 = 0.
Solution
0 + 0 = 0 (Identity)
⇒ a × (0 + 0) = a × 0
⇒a×0+a×0=a×0
(Multiplying both sides by a)
(Distributive property)
a × 0 + a × 0 + (–a × 0) = a × 0 + (–a × 0)
a × 0 + [a × 0 + (–a × 0)] = a × 0 + (–a × 0)
a×0+0=0
∴a×0=0
R
(Associative Property)
(Additive Inverse)
(Identity)
Wo rke d Exa m p le 1 . 7
R
Worke d E x a m p l e 1. 8
Use the properties of the integers to prove that
for all integers a and b, (a)(–b) = –ab.
Use the properties of the integers to prove that for
all integers a and b, (–a)(–b) = ab.
Solution
Solution
(a)(–b) + (a)(b) = a(–b + b) (Distributive
Property)
= a(0)
(–a)(–b) + (–a)(b) = –a(–b + b) (Distributive
Property)
(Additive Inverse)
= –a(0)
(Additive Inverse)
∴ (a)(–b) + (a)(b) = 0
⇒ (–a)(–b) + (–a)(b) = 0
Therefore, (a)(–b) is the additive inverse
of (a)(b).
Therefore, (–a)(–b) is the additive inverse of (–a)(b).
But (a)(b) = ab and its additive inverse is –ab.
∴ (a)(–b) = –ab
8
( Adding –a × 0 to both sides)
But (–a)(b) = –ab and its additive inverse is ab.
∴ (–a)(–b) = ab
R
1
Worke d Exa m p le 1 . 9
REAL NUMBERS
(i) Find the sum of the squares of the four consecutive integers, –2, –1, 0 and 1.
(ii) Show that the sum of the squares of any four consecutive integers is always an even number.
Solution
(i) (–2)2 + (–1)2 + (0)2 + (1)2 = 4 + 1 + 0 + 1
=6
(ii) n, n + 1, n + 2, and n + 3, n ∈ Z, are four consecutive integers.
∴ (n)2 +(n + 1)2 +(n + 2)2 +(n + 3)2 = n2 + n2 + 2n +1 + n2 + 4n + 4 + n2 + 6n + 9
= 4n2 + 12n + 14
= 2(2n2 + 6n + 7) [an even number as 2 is a factor]
∴ The sum of any four consecutive integers is an even number.
Exercise 1.2
5. Use the properties of the integers to prove
that –5 × 8 = –(5 × 8).
1. Evaluate each of the following:
(i) –8 – 5 + 13
(iii) (–5)(–8)
(ii) (2)(–3)
(iv) 3(4)2 + 2(4) – 56
6. A box contains oranges and
grapes. An equal number of
the oranges and grapes are
rotten. _32 of all the oranges
are rotten and _43 of all the
grapes are rotten.
What fraction of the total number
of pieces of fruit in the box are rotten?
3(5 – 2)2 – 3(4 – 2)2 + 5(3)3
(v) ________________________
(5 – 2)2
2. Evaluate each of the following, leaving your
answers in their simplest form:
3
2 2 5
1__ × 1__ + __ × __
4
3 3 6
14 __
___
________________
3
×
(vi)
(i)
1
15 8
7__ – 4
2
8.4(19.6 – 12.2)2
1
(ii) 2__ × 5
(vii) _______________
2
(14.4 – 12.2)3
3
__
2
__
(iii) 1 × 1
4
3
3 ___
13
__
–
5 20
______
(viii)
5
__
8
1
__
1
__
(ix) 3(2.5 – 1.2)2
–
2 8
2 5
(v) __ + __
3 6
(iv)
7. Alice and Bob share an allotment. The ratio of
the area of Alice’s portion to the area of Bob’s
portion is 3 : 2. They each grow vegetables
and fruit on the allotment.The entire allotment
is covered by vegetables and fruit in the ratio
7 : 3. On Alice’s portion of the allotment, the
ratio of vegetables to fruit is 4 : 1. What is the
ratio of vegetables to fruit in Bob’s portion?
8. Three cans of juice fill _23 of a 1 litre jug.
How many cans of juice are are needed to
completely fill eight 1 litre jugs?
3. Use the properties of the integers to prove
that 4 × –5 = –(4 × 5).
9. A rectangle has a length of _35 units and an area
of _13 units2. What is the width of the rectangle?
4. Use the properties of the integers to prove
that –3 × –6 = (3 × 6).
10. In the diagram, the number line is marked at consecutive integers, but the numbers themselves are
not shown. The four large red dots represent two numbers that are multiples of 3 and two numbers
that are multiples of 5. Which of the labelled points (black dots) represent a number that is a multiple
of 15? Give an explanation for your choice.
+
A
+
B
+
+
P
+
J
+
Q
+
G
+
R
+
+
C
+
+
D
+
+
+
9
REAL NUMBERS
1
11. The integers from 1 to 9 are listed on a whiteboard. If an additional m 8s and n 9s are added to the
list, then the average of all the numbers is 7.3. Find m + n.
12. Five square tiles are shown. Each tile has a side of integer length. The side lengths can be arranged
as consecutive integers. The sum of the areas of the five squares is 1,815.
(i) Show that the sum of the
squares of five consecutive
integers is divisible by 5.
(ii) Find the dimensions of
the largest square.
14. Seán has a pile of tiles, each measuring 1 cm by 1 cm. He tries to put these small tiles together to
form a larger square of length n cm, but finds that he has 92 tiles left over. If he had increased the
side length to (n + 2) cm, then he would have been 100 tiles short. How many tiles does Seán have?
15. A palindromic number is a positive integer that is the same when read forwards or backwards.
For example, 31213 and 1237321 are palindromic numbers.
(i) Find the total number of three digit palindromic numbers.
(ii) Determine the total number of palindromic numbers between 106 and 107.
(iii) If the palindromic numbers in part (ii) are written in order, find the 2,125th number on the list.
1.3 IRR ATIONAL NUM BERS
In the right-angled triangle shown, the value for x can be found
using the theorem of Pythagoras. Here is the solution:
x2
12
=
x2
=1+1
x2
=2
+
__
x = √2
__
12
1
x
1
(as x > 0)
Can √ 2 be written as a fraction? This problem preoccupied the ancient
Greek mathematicians for many
__ years. Around 500 BC, Hippasus, a follower
of Pythagoras, proved that √ 2 could not be written as a fraction. Pythagoras,
who did not believe in the existence of irrational numbers, was so enraged by
this proof that he had Hippasus thrown overboard from a ship and Hippasus
subsequently drowned.
__ Numbers that cannot be written as fractions are called
irrational numbers. √ 2 was the first known irrational number.
Hippasus, a follower
of Pythagoras
__
While √ 2 cannot be written__as a fraction, it is possible to
find an approximation
for √ 2 . A calculator gives the
__
approximation √ 2 = 1.414213562, but this decimal
goes on forever with no pattern or repetition.
10
1
__
Proof that
√ 3 is Irrational
__
To prove: √ 3 is irrational.
__
Proof: Assume that √ 3 is rational and can therefore be written in the form _ba , a, b ∈ Z, b ≠ 0.
Also, assume that the fraction _ba is written in simplest terms i.e. HCF(a, b) = 1.
__
√3 = _ba ,
2
a
⇒ 3 = __
2 (squaring both sides)
b
∴ a2 = 3b2
(*)
REAL NUMBERS
The proof of this result is another example of proof by contradiction.
As b2 is an integer, a2 has to be a multiple of 3, which means that 3 divides a2.
If 3 divides a2, then 3 divides a. (Worked Example 1.3)
∴ a = 3k, for some integer k. Substituting 3k for a in (*) gives,
(3k)2 = 3b2
9k2 = 3b2
⇒ b2 = 3k2
As k2 is an integer, b2 has to be a multiple of 3, which means that 3 divides b2.
Therefore, 3 divides b. If 3 divides a and 3 divides b, then this contradicts the assumption that
HCF(a, b) = 1. This completes the proof.
__
__
Constructing
√ 2 and √ 3
__
__
√2 and √3 cannot be written as fractions, but can be constructed.
__
Construct √ 2
1. Let the line segment AB be of length 1 unit.
4. Mark the intersection, C, of the circle and m.
C
m
B
A
2. Construct a line m perpendicular to [AB] at B.
m
A
B
B
A
5. Draw the line
__ segment CA.
|AC| = √ 2
C
m
3. Construct a circle with centre B and radius [AB].
m
A
A
B
B
11
1
Proof:
|AB| = |BC| = 1
(radii of circle)
REAL NUMBERS
|AB|2 + |BC|2 = |AC|2 (Theorem of Pythagoras)
12 + 12 = |AC|2
|AC|2 = 2
__
∴ |AC| = √ 2
__
Construct √ 3
1. Let the line segment AB be of length 1 unit.
B
A
5. Draw the line
__ segment [CD].
|CD| = √ 3
C
2. Construct a circle with centre A and radius
length |AB|.
E
B
A
B
A
D
Proof:
3. Construct a circle with centre B and radius
length |AB|.
CD is the perpendicular bisector of [AB]
(Construction).
1
∴ |AE| = |EB| = __
2
|AC| = |BC| = 1 (Construction)
|AE|2 + |EC|2 = |AC|2 (Theorem of
Pythagoras)
( 12 ) + |EC| = 1
__ 2
2
2
1 3
|EC|2 = 1 – __ = __
4 4
B
A
__
__
∴ |EC| =
√4 = 2
√3
___
3
__
|CD| = 2 |EC|
__
=2
4. Mark the intersection of the two circles as C
and D.
C
B
A
D
12
( )
√3
___
__
⇒ |CD| = √ 3
2
1
Exercise 1.3
__
1
2
1
90º
1.5 90º
1
1
90º
1
0.5
0
–1.5 –1 –0.5
0 0.5
–0.5
90º
1
1.5
2
2.5
REAL NUMBERS
1. Copy the diagram and use it to show √ 5 on the numberline.
__
2. Using a diagram similar to that used in Question 1, show √ 7 on the numberline.
__
3. Prove by contradiction that √ 2 is irrational.
4. p, the ratio of the circumference of a circle to its diameter, is also an irrational number. The British
mathematician John Wallis (1616–1703) discovered the following formula for approximating p:
(2n)2
22
62 × ... × ______________
42 × _____
= _____ × _____
2 (1)(3) (3)(5) (5)(7)
(2n – 1)(2n + 1)
p
__
(i) Use the first seven terms of the product to approximate p to four decimal places.
(ii) Use your calculator to approximate p to four decimal places.
(iii) Assuming that your calculator has given the true approximation to four decimal places, calculate
the percentage error in using seven terms of Wallis’ formula (answer to two decimal places).
________
__
_______
__
5. Show that x = √ 3 + 2√ 2 – √ 3 – 2√ 2 is a rational number. ( Hint: find x2. )
6. Two quantities are in the golden ratio, if the ratio of the sum of the quantities to the larger quantity
is equal to the ratio of the larger quantity to the smaller one. In the rectangle below a > b and
a + b __
a
_____
a = b ; therefore, the ratio a : b is the golden ratio.
This ratio is a constant and also irrational. It can be shown that
a
b
the ratio is the positive solution to the equation:
1
r = 1 + __r
a
(i) Solve the equation to find the golden ratio.
a+b
(ii) Use your calculator to approximate the ratio to four
decimal places.
1.4 ROUNDING AND SIGNIFICANT FIGURES
Rounding to Decimal Places
In geometry, the number of times the diameter of a circle divides into the circumference is called p.
We normally substitute 3.14 for p in these calculations. However, 3.14 is just an approximation:
„ There are infinitely many decimal places in p.
„ p is 3.141592654 to nine decimal places.
„ For simplicity, we often write p as 3.14, i.e. to two decimal places.
„ Engineers use 3.1416 as an approximation for p.
13
R
Wo rke d E
Exa
x a m p le
le 1.10
Write the following correct to one decimal place:
REAL NUMBERS
1
(i) 2.57
(ii) 39.32
Solution
(i) 2.57
When rounding to one decimal place, we look at the second number after the decimal
point. If this number is 5 or greater, we round up to 2.6. Otherwise, the corrected answer is
2.5. As 7 is the second number after the decimal point, we round up to 2.6.
Answer = 2.6
(ii) 39.32
Here, the second number after the decimal point is 2, which is less than 5. Therefore, the
number rounded to one decimal place is 39.3.
Answer = 39.3
Significant Figures
R
Wo rke d E
Exa
x a m p le
le 1.11
Correct the following numbers to two significant figures:
(i) 3.67765
(ii) 61,343
(iii) 0.00356
Solution
(i) 3.67765
The first significant figure in a number is the
first non-zero digit in the number. In this number,
3 is the first significant figure in the number.
We need to correct to two significant figures, so we
look at the third significant digit. If this number
is 5 or greater, we round up the second digit.
The third digit is 7, so the corrected number is 3.7.
14
3.6|7765
1st sig. figure
2nd sig. figure
Answer = 3.7
(ii) 61,343
61,|343
1st sig. figure
2nd sig. figure
Answer = 61,000
Note that all other digits after the rounded digit change
to zero.
(iii) 0.00356
0.0035|6
The third significant digit is 6. Therefore, the
rounded number is 0.0036.
1st sig. figure
2nd sig. figure
Answer = 0.0036
REAL NUMBERS
Here, the third digit is 3, which is less than 5.
Therefore, the rounded number is 61,000.
1
1.5 ORDERS OF MAGNITUDE
AND SCIENTIFIC NOTATION
When doing calculations, scientists often use very large numbers or very small numbers. For example,
the speed of light is about 300,000,000 metres per second.
Very large or very small numbers can be awkward to write down. So, scientists use scientific notation to
write down these numbers.
R
Wo rke d Exa m p le 1. 1 2
Write the following numbers in scientific notation:
(i) 725,000,000,000
(iii) 0.0000056
(ii) 980,000
(iv) 0.000000034
Solution
(i) First, note that dividing a number N by 10n, where n ∈ N, moves the decimal point n places to
the left.
144.25
= 1.4425 (Decimal point moves two places to the left)
For example, _______
102
725,000,000,000
725,000,000,000 = _______________
× 1011
1011
= 7.25 × 1011
15
1
980,000
(ii) 980,000 = ________
× 105
105
REAL NUMBERS
= 9.8 × 105
(iii) Note that dividing a number N by 10n, where n is a negative integer, moves the decimal n places
to the right.
0.00146
1
For example, ________
= 0.00146 × ____
10–3
10–3
= 0.00146 × 103 (Rules of indices)
= 1.46
(Decimal point moves three places to the right)
0.0000056
0.0000056 = __________
× 10–6
10–6
= 5.6 × 10–6
0.000000034
(iv) 0.000000034 = ____________
× 10–8
10–8
= 3.4 × 10–8
Orders of magnitude
Orders of magnitude are generally used to make very approximate comparisons. If two numbers differ
by one order of magnitude, one is about ten times larger than the other.
R
Wo rke d Exa m p le 1 . 1 3
By how many orders of magnitude does 345,632 differ from 567,123,423?
Solution
Write both numbers in scientific notation:
345,632 = 3.45632 × 105
567,123,423 = 5.67123423 × 108
≈ 1 × 105
≈ 10 × 108
= 100 × 105
= 101 × 108
= 105
= 109
109
____
= 109 – 5 = 104
105
Therefore, both numbers differ by four orders of magnitude.
16
1
Exercise 1.4
5. Calculate each of the following, giving your
answers as decimal numbers:
(i) 5.1456
(iv) 62.1235321
(i) 3.4 × 103 + 2.8 × 10 3
(ii) 7.2983
(v) 23.7654
(ii) 5.2 × 109 + 3.5 × 10 9
(iii) 17.8943
(vi) 0.07893
(b) Write these numbers correct to two
decimal places:
6. The following numbers are written in scientific
notation. Rewrite the numbers in ordinary
form.
(i) 1.263
(iv) 21.3
(i) 2 × 106
(iv) 6.47 × 105
(ii) 5.9876
(v) 22
(ii) 1.69 × 104
(v) 6.12 × 101
(iii) 21.456
(vi) 0.00391
(iii) 2.48 × 103
(vi) 9.43 × 10 5
2. (a) Write these numbers correct to two
significant figures:
7. Write the following in the form a × 10n,
where 1 ≤ a < 10, a ∈ R, n ∈ Z:
(i) 0.00985
(vi) 0.000849
(i) 0.000036
(iv) 0.00063
(ii) 0.00234
(vii) 0.238
(ii) 0.0005613
(v) 0.0078
(iii) 0.0125
(viii) 52.00285
(iii) 0.0345
(vi) 0.0011
(iv) 0.000000785
(ix) 52.487
(v) 1.000034
(x) 967,333
(b) Write these numbers correct to one
significant figure:
(i) 32.14
(iv) 1,698
(ii) 3.857
(v) 5,965
(iii) 19,345
(vi) 999
3. Write these numbers in scientific notation:
(i) 34,000,000
(vi) 0.000032
(ii) 0.25
(vii) 5,000,000
(iii) 4,570
(viii) 0.6464
(iv) 0.0001258
(ix) 532,600
(v) 7,206
(x) 5,000
4. Write these as decimal numbers:
REAL NUMBERS
1. (a) Write these numbers correct to three
decimal places:
8. Write the following numbers in the form
a × 10n, where 1 ≤ a < 10:
(i) 0.00068
(iv) 0.0000000097
(ii) 0.0000328
(v) 0.00000056
(iii) 0.0657
(vi) 0.0030307
9. The following numbers are written in scientific
notation. Rewrite the numbers in ordinary
form.
(i) 1.5 × 10 –3
(iii) 3.5 × 10 –5
(ii) 2.54 × 10 –4
(iv) 6.67 × 10 –6
10. By how many orders of magnitude do the
following numbers differ:
(i) 868, 932, 145 and 284
(ii) 453, 987, 312 and 3548
(iii) 767, 894, 567,000 and 23,000,000
(i) 2.65 × 102
(vi) 4 × 10 –2
(iv) 0.1 and 0.00042
(ii) 4.53 × 10 –3
(vii) 2.64 × 107
(v) 1.8 and 234
(iii) 7.2 × 106
(viii) 7.612 × 103
(iv) 1.7 × 10 –5
(ix) 2.76 × 108
(v) 3 × 10 2
(x) 3.02 × 10 –9
17
REAL NUMBERS
1
Revision Exercises
1. The German mathematician Christian
Goldbach conjectured that every odd positive
integer greater than 5 is the sum of three
primes. Verify this conjecture for each of the
following odd integers:
(i) 11
(ii) 33
(iii) 97
(iv) 199
9. Write the following numbers correct to two
significant places:
(v) 17
(iv) 0.000054
(ii) 0.134
(v) 652,494
(iii) 2.00062
(vi) 0.000814
2. Two bikers are riding in a circular path.
The first rider completes a circuit in 12 minutes.
The second rider completes a circuit in
18 minutes. They both started at the same
place and at the same time and go in the same
direction. After how many minutes will they
meet again at the starting point?
10. Given a line segment of length one unit, show
clearly how
__ to construct a line segment of
length √ 2 units.
3. Use the properties of integers to prove the
following:
12. (a)
(i) 5 × –3 = –15
11. Given a line segment of length one unit, show
clearly how
__ to construct a line segment of
length √3 units.
(iii) Using your answers to parts (i) and
(ii), find the HCF and LCM of 1,768
and 3,192.
5. (i) List the smallest five consecutive integers
that are composite.
(b) Evaluate the following, giving your answer
in scientific notation:
3
2
1
3__ × 2__ + 1___
7
5
10
_______________
3 × 104
(ii) Find 10,000 consecutive integers that are
composite.
6. Express each of the following numbers as the
product of prime factors, and hence, find the
LCM and HCF of each pair:
(i) 68 and 102
(iii) 104 and 351
(ii) 69 and 123
(iv) 123 and 615
7. (a) What fraction when added to _14 gives _13?
(b) A mathematician states that her children’s
ages are all prime numbers that multiply
together to give 7,429. She also says that
two of her children are teenagers.
(i) How many children does she have?
(ii) What are their ages?
8. A palindromic number is a number that reads
the same forwards and backwards. For example,
52,325 is a palindromic number. All four-digit
palindromic numbers have 11 as a prime factor.
(i) Find the prime factorisations of the
palindromic numbers 2,332 and 6,776.
(ii) Hence, find the HCF and LCM of 2,332
and 6,776.
(i) Find the values of the primes p and
q, if p3 × 13 × q = 1,768.
(ii) Find the values of the primes m and n,
if 24 × m × n = 3,192.
(ii) –5 × –3 = 15
4. Show that no integer of the form n3 + 1 is
prime, other than 2 = 13 + 1.
18
(i) 852,233
(
13.
)
(i) P represents a pointer on a gauge.
What is the decimal value shown by the
pointer?
P
0
0.01
(ii) A computer shop buys a batch of iPod
nanos for €99 each and marks the price
up by _13. The goods fail to sell so they are
included in the next sale where all prices
are reduced by _14.
What price would you pay for an iPod
nano in the sale?
(iii) John calculates correctly 85 × 142 = 12,070.
Using this information, what should his
answer be for 12,070 ÷ 850?
__
14. Prove that √5 is irrational.
16. A man died, leaving some money to be
divided among his children in the following
manner:
1
of what
„ €x to the first child plus __
16
remains.
1
„ €2x to the second child plus __
of what
16
then remains.
1
„ €3x to the third child plus __
of what then
16
remains and so on.
1
REAL NUMBERS
15. (i) A single-celled protozoa is about one
tenth of a millimetre in diameter. Write
down its diameter in metres, giving your
answer in scientific notation.
When all the money was distributed, each
child received the same amount of money
and no money was left over.
(ii) A bag contains 350 disks. Seán takes
four-fifths of the disks out of the bag and
divides them into seven equal groups.
How many disks are in each group?
How many children did the man have?
17. The digits 1, 2, 3, 4 and 5 are each used once to create a five-digit number vwxyz, which satisfies
the following conditions:
„ The three digit number vwx is odd.
„ The three digit number wxy is divisible by 5.
„ The three digit number xyz is divisible by 3.
Determine the six five-digit numbers that satisfy all three conditions.
18. A well known series for the irrational p is:
p
__
2
1 1×2 1×2×3 1×2×3×4
= 1 + __ + ______ + _________ + _____________ + ...
3 3×5 3×5×7 3×5×7×9
By summing the first 10 terms of the series, find an approximation for p to two decimal places.
19
chapter
h
Algebra I
Learning Outcomes
In this chapter you will learn to:
Add, subtract, multiply and divide algebraic terms
Factorise quadratic expressions
Factorise certain cubic expressions
Divide algebraic expressions using long division
2
2 .1 E XPRESSIONS
2.1
Algebra has many uses, from the design
of computer games to the modelling
of weather patterns.
To be able to use algebra, we must first
understand the rules involved in the
basic operations of adding, subtracting,
multiplying and dividing algebraic terms and expressions.
Notation in Algebra
ALGEBRA I
In 5x, the variable is x.
In 20y, the variable is y.
In 5x, the coefficient is 5.
In 20y, the coefficient is 20.
In x, the coefficient is 1.
In 10x + 2, the constant is 2.
12(x) is an example of a term.
This is written as 12x.
x + 5y – 7 contains three terms:
x (a variable)
5y (a constant times a variable)
–7 (a constant)
.
5x + 2 is an expression. Other examples of
expressions include x + 3y, 8y2 and 4pr3 – 7.
A polynomial in x has the form anxn + an – 1xn – 1 + an – 2xn – 2 + ... +a2x2 + a1x1 + a0, where all the
coefficients (a0, a1, a2, ..., an – 1, an) are constants and the powers are non-negative whole numbers.
21
2
The degree (or order) of the polynomial is equal to the highest power of x.
A polynomial of degree 1 is called linear.
A polynomial of degree 3 is called cubic.
A polynomial of degree 2 is called quadratic.
A polynomial of degree 4 is called quartic.
For example, 5x6 + 2x2 – 3x + 7 is a polynomial in x of degree 6. It has four terms.
The coefficient of x is –3 and the constant is 7.
Addition/Subtraction
When adding or subtracting algebraic terms, we must always remember the following rules:
ALGEBRA I
Examples:
5x + 6y + 4x – 3y = 9x + 3y
p + 2q + 3p – 4q = 4p – 2q
Examples:
20y2 + 8y2 = 28y2
xy – 3w2 + 5xy + 2w2 = 6xy – w2
Multiplying
Unlike when adding or subtracting, in algebra any term may be multiplied by any other term.
When we multiply terms, we encounter another set of rules that are important to understand.
Example: (4x)(5y)
= (4)(x)(5)(y)
= (4)(5)(x)(y)... (Commutative)
= 20xy
Example: (4x3)(3x5)
= (4)(x)(x)(x)(3)(x)(x)(x)(x)(x)
= (4)(3)(x)(x)(x)(x)(x)(x)(x)(x)... (Commutative)
= 12x8
22
2
Important Products
Worke d Exa m p le 2 . 1
Expand and simplify:
(i) 4(2a – b)(a – 5b)
(ii) 23x – x(2x + 3)2
(iii) (4x – 5)3
(i) 4(2a − b)(a − 5b)
(i
(ii) 23x – x(2x + 3)2
(iii) (4x – 5)3
(iii
4[(2a(a – 5b) – b(a – 5b)]
23x – x[(2x)2 + 2(2x)(3) + (3)2]
(4x)3 – 3(4x)2(5) + 3(4x)(5)2 – (5)3
4[2a2 – 10ab – ab + 5b2]
23x – x[4x2 + 12x + 9]
64x3 – 240x2 + 300x – 125
4[2a2 – 11ab + 5b2]
23x – 4x3 – 12x2 – 9x
8a 2 – 44ab + 20b2
–4x3 – 12x2 + 14x
ALGEBRA I
Solution
Pascal’s Triangle
(a + b)5 would take a long time to expand and simplify using the methods
employed above. However, by looking at the expansion of (a + b)n we can
spot a pattern that we may be able to use for other expressions (n ∈ N or n = 0).
(a + b)0 = 1
(a + b)1 = a1 + b1
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
If we examine the powers, we notice that, for n ≥ 1, the power of each term in the expansion add up to
the power of the original expression.
(a + b)0= 1
(a + b)1 = a1 + b1
(a + b)2 = a2 + 2a1b1 + b2
(a + b)3 = a3 + 3a2b1 + 3a1b2 + b3
(a + b)4 = a4 + 4a3b1 + 6a2b2 + 4a1b3 + b4
(a + b)5 = a5 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + b5
23
2
If we highlight (a + b)5:
(a + b)5 = a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + a0b5
we notice that the first term is a5 and then the powers of a decrease by 1 each term and eventually reach
zero. Also, the powers of the b start at zero and increase by 1 each term.
Examining the coefficients of each expansion,
we notice another pattern.
1
1
The next row of coefficients can be calculated by
adding the pairs of coefficients from the row above.
ALGEBRA I
This pattern was first described by the
French mathematician Blaise Pascal (1623–1662)
and is named in his honour.
1
1
1
1
1
2
3
4
1
1
3
6
1
2
4
1
1
1
1
3
4
5
10
There are n + 1 terms.
The coefficients of the first two terms are 1 and n.
The sum of the powers of x and y in each term is n.
Use Pascal’s triangle to expand (x – 2)3.
Solution
From Pascal’s triangle we can determine the initial coefficients as:
1 3 3 1
We next fill in the powers of each term.
The x term will start with a power of 3 and decrease to zero.
The (–2) term will start with a power of zero and increase to 3.
∴ (x – 2)3 = (x + (–2))3 = 1x3 (–2)0 + 3x2(–2)1+ 3x1 (–2)2 + 1x0(–2)3 = x3 – 6x2+ 12x – 8
Wo rke d Exa m p le 2 . 3
Use Pascal’s triangle to expand (2p + 3)4.
Solution
From Pascal’s triangle we can determine the coefficients as:
1 4 6 4 1
The 2p term will start with a power of 4 and decrease to zero.
The 3 term will start with a power of zero and increase to 4.
∴ (2p + 3)4 = 1(2p)4 (3)0 + 4(2p)3 (3)1+ 6(2p)2 (3)2 + 4(2p)1 (3)3 + 1(2p)0(3)4
3
6
Wo rke d Exa m p le 2 . 2
24
1
1
In general, using Pascal’s triangle in the expansion of (x + y)n:
= 16p4 + 96p3 + 216p2 + 216p + 81
1
1
4
10
1
5
1
2
Exercise 2.1
1. If p = 3, q = – 4 and r = 7, find the value of:
(i) 2pq
(iv) pqr – q2
(ii) (q + r)p
(v)
q+p
(iii) ______
2r
√
3
___________
2
q + rp + r
__________
–q
___
p
6. Expand:
(i) (3p – 4)(b2 + 4b + c)
(ii) (b2 + 4b + c)(4y – x)
(iii) (3a – 1)(a + 3b)(2a + b)
(iv) 5(x + y)(x + 4y)(y + x)
(v) (z + x)2(z – x)2
2. Expand and simplify:
(i) 3x(x + 4) + 5(3x – 2)
7. Expand:
(i) (p + q)2
(vi) (y – 3)3
(iii) 12x(3x2 + 2x + 1) + 5x(2x – 4)
(ii) (p – q)2
(vii) (x + 4)3
(iv) –y(2y – 3x) – y(xy + 3x)
(iii) (p – q)(p + q)
(viii) (2a + 5b)3
(v) b(b2 + 4b + c) – 4c(a2 + b)
(iv) (x – 3)(x + 3)
(ix) (p – 8q)3
(v) 2x(4x + 3)2
(x) (x + y)(x – y)y2
3. Expand each of the following expressions.
For each polynomial, state:
8. Expand:
(i) (2p – q)(2s + 4r)
(v) 5(2a + 9b)3
(b) The value of the constant term
(ii) (a – 3b)(a + 3b)
(vi) (x2 – y2)(x – y)
(c) The number of terms
(iii) (4m + 7n)3
(vii) [(a – b)b]3
(d) The coefficient of the x term
(iv) (9x – 2y)3
(a) The degree
(i) (x + 2)(x – 3)
9. Using Pascal’s triangle, expand the following
polynomials:
(ii) (2x – 5)(3x + 4)
(iii) –x(5x2 – 3)(2x2 – 3)
(iv) (x3 – 4)(2x2 – 8x + 3)
(v)
(11x4
–
3x3
+
x2
ALGEBRA I
(ii) 3(a2 – b) – (a – 3b)
– 3x + 1)(4x – 7)
4. Multiply out the following expressions:
(i) (5x + 1)(7x + 1)
(ii) (x + 3)(4x – 3)
(iii) 5(3s – t) (2t – 1)
(iv) (x – y) (2x + 5y)(3x)
(v) (xy – y) (y – 6xy)(y)
5. Expand:
(i) (3p + 4)2
(iv) (a + b)(x + 5)2
(ii) (4y – x)2
(v) (y + x)3(2y – 3x)
(iii) 2(3a – 1)2
(vi) 3p(p + q)(p – q)
(i) (a + 1)4
(iv) (2a + 3b)3
(ii) (b – 3)3
(v) (3y – 4x)4
(iii) (x + y)5
(vi) (3x – 2y)5
1
10. In a maths competition, a students gets
8 marks for a correct answer, and 3 marks are
deducted if the answer is incorrect. One mark
is awarded for any question not attempted.
If there are 20 questions on the test, write
an expression for a student’s total mark if the
student got x questions correct from a total of
y questions attempted.
1
11. Mark is y years old. In five years time, Aoife
will be twice as old as Mark and Daniel will
be two years younger than Aoife is now.
Write an expression in y for the sum of their
three ages now.
25
2
12. Express as a polynomial the area and
perimeter of each of the following
shapes:
(i)
13.
1 An open-topped tank is in the shape of a
cuboid. The length of the tank is (x + 7) cm.
Its breadth is 3 cm less than its length and its
height is 1 cm more than its breadth. Write an
expression in terms of x for the volume and
surface area of the tank.
x+2
(ii)
ALGEBRA I
2x – 3y
14.
1 Find the value of the coefficient of x3 in the
expansion of (x + y)7.
x+y
15.
1 Find the value of the third term in the
expansion of (4a – 5b)8.
1 The middle term of the expansion of
16.
(ax + ay)n is 14,580x3y3. Find the value of
a and the value of n, where a, n ∈ N.
2.2 FACTORISING
Another important skill is that of factorising expressions.
Highest Common Factor
x2 – 3x = x(x – 3)
Difference of Two Squares
4x2 – 25y2
= (2x)2 – (5y)2 (writing each term as a square)
= (2x – 5y)(2x + 5y)
Quadratic Trinomials
2x2 + 13x + 15 = (2x + 3)(x + 5)
Wo rke d Exa m p le 2 . 4
Factorise: (i) 49x2 – 225y2
(ii) 3x2 + 11x – 20
Solution
(i) 49x2 – 225y2
Write each term as a square.
(7x)2 – (15y)2
(7x – 15y)(7x + 15y)
26
(iii) 8x2 – 18x + 9
2
(ii) 3x2 + 11x – 20
If the constant term is negative, then one factor of 20 is positive and the other is negative
3x
+
5
3x
+
4
3x
–
4
x
–
4
x
–
5
x
+
5
3x × x = 3x2
3x × x = 3x2
5 × –4 = –20
We now use the arrows:
4 × –5 = –20
We now use the arrows:
–4 × 5 = –20
We now use the arrows:
3x × –4 = –12x
3x × –5 = –15x
3x × 5 = +15x
x × 5 = +5x
–7x
x × 4 = +4x
–11x
x × –4 = –4x
+11x
We must be on the right
track, as we have the right
number but the wrong sign.
So (3x – 4)(x + 5) are the correct factors.
ALGEBRA I
3x × x = 3x2
(iii) We can also use the Guide Number Method to factorise quadratic trinomials.
8x2 – 18x + 9
Step 1 Multiply the coefficient of x2 by the constant.
8x2 – 18x + 9
8 × 9 = 72
Step 2 Find two factors of 72 that add to give the coefficient of the middle term, i.e. –18.
–12 and –6
Step 3 Use the answers from Step 2 to rewrite 8x2 – 18x + 9 as follows:
8x2 – 12x – 6x + 9
4x(2x – 3) – 3(2x – 3)
(Factorise by grouping)
(4x – 3)(2x – 3)
(Distributive property)
The factors of 8x2 – 18x + 9 are (4x – 3)(2x – 3).
Sum and Dif ference of Two Cubes
We may also be asked to factorise certain cubic expressions.
27
2
Wo rke d Exa m p le 2 . 5
Factorise:
(i) x3 + 125
(ii) 64p3 – 27q3
(iii) 1 + 216w3
Solution
(i) x3 + 125
(iii) 1 + 216w2
(ii
Write each term as a cube.
Write each term as a cube.
(x)3 + (5)3
(1)3 + (6w)3
= (x + 5)(x2 – 5x + 25)
= (1 + 6w)[ (1)2 – (1)(6w) + (6w)2 ]
= (1 + 6w)(1 – 6w + 36w2)
ALGEBRA I
(ii) 64p3 – 27q3
Write each term as a cube:
(4p)3 – (3q)3
= (4p – 3q)[(4p)2 + (4p)(3q) + (3q)2]
= (4p – 3q)(16p2 + 12pq + 9q2)
Wo rke d Exa m p le 2 . 6
Factorise fully:
(i) 9px2 + 24px – 9p
(ii) 5x3 – 625y3
Solution
(ii) 5x3 – 625y3
(i) 9px2 + 24px – 9p
First find the highest common factor:
First, find the highest common factor:
3p[3x2 + 8x – 3]
5[x3 – 125y3]
Now factorise the quadratic:
Now factorise the expression inside the
brackets:
3p(3x – 1)(x + 3)
5(x – 5y)(x2 + 5xy + 25y2)
Exercise 2.2
Factorise fully:
28
1. 4ab2 – 12ab3
4
4. 5x2 + 12x + 4
7. 2y2 + 11y – 63
2. 7x2 + 9x + 2
5
5. x2 – 3x – 18
8. 7x2 + 2x – 57
3. 3y2 – 4y – 7
6
6. 3x2 + 10x – 8
9. 25a2 – 1
10. 2x2 – 9x + 4
1
17. 9x2 – 21x + 10
2 12x2 – 18x –12
24.
11. 5x2 + 52x + 96
1
18. 4y2 + 23y + 19
2
25. 48y3 + 48y2 + 12y
12. 64a2 – 81b2
1
19. 6x2 + 37x + 45
2
26. 16x2 –100
13. 2x2 – 23x – 12
20. 36p2 – 100q2
2
27.
2 2x3 – 8x
14. 4x2 + 7x + 3
21.
2 12x2 + 11x – 56
28.
2 6pq2 + 11pq +4p
15. 10y2 + 27y + 17
2
22. 8x2 – 22x + 15
2
29. x4 – 36x2
16. 9x2 + 3x – 2
2
23. 10x2 – 4x – 6
2
Factorise fully the following expressions:
34.
3 8a3 + 27b3
38.
3 3x3 – 243
31. p3 + 8
35.
3 125p3 + 512q3
39.
3 128 + 16x3
32. x3 – y3
36.
3 1,000x3 – 729
40.
4 54a4 + 432ab3
33. 64a3 – 1
37. 343c3 + d3
3
ALGEBRA I
30. x3 – 27
Factorise fully the following expressions:
41. (x + 2)2 – (x + 3)2
45.
4 a2b2 – 2ab + 1
49
49. a4 + a
42. x2 + 2px + p2
46.
4 x4 – y4
50.
50 a2 – (b + c)2
43. a2c2 – b2
4
47. (x – y)2 – 9
51.
5 ab5 – ab2
44. x4 – 25
4
48. 8x2 – 18xy + 9y2
52.
52 x4 – 7x2 – 18
2 .3 ALGEBR AIC FR ACTIONS I:
2.3
ADDITION AND SUBT R ACTION
When adding or subtracting two algebraic fractions, we first find the lowest common denominator (LCD).
Worke d Exa m p le 2 . 7
2x + 5 x – 1
Write as a single fraction: ______ – _____ – 2
2
9
Solution
2x + 5
______
2
x – 1 2 9(2x + 5) – 2(x – 1) – 18(2)
– _____ – __ = _______________________
9
1
18
18x + 45 – 2x + 2 – 36 _________
16x + 11
_____________________
=
=
18
18
29
2
Wo rke d Exa m p le 2 . 8
(i) Write as a single fraction:
5
2
______
– ______
3x – 5 4x – 1
b
1
1
(ii) Write _____ – _____ as a single fraction in the form _____
, where a, b ∈ N.
x–2 x+2
xa – b
Solution
5
2
(i) ______ – ______
3x – 5 4x – 1
5(4x – 1) – 2(3x – 5)
= _________________
(3x – 5)(4x – 1)
ALGEBRA I
20x – 5 – 6x + 10
= ________________
(3x – 5)(4x – 1)
14x + 5
= _____________
(3x – 5)(4x – 1)
1(x + 2) – 1(x – 2)
1
1
(ii) _____ – _____ = ________________
x–2 x+2
(x – 2)(x + 2)
x+2–x+2
_____________
=
x2 – 4
4
= _____
x2 – 4
Wo rke d Exa
Worke
E x a m p le
le 2.9
5x – 10
.
Simplify _______
x2 – 4
3a2 – 6a
Simplify ________.
2–a
Solution
Solution
We factorise both numerator and
denominator.
5x – 10 = 5(x – 2)
x2 – 4 = (x – 2)(x + 2)
5(x – 2) 1
5x – 10 ____________
=
∴ _______
2
(x – 2)(x + 2)
x –4
5
_____
=
x + 21
Wo rke d Exa m p le 2 . 1 1
(i) Express as a single fraction in its simplest form:
3x – 1
4
2
_____
– __________ + _____
x – 1 x2 – 4x + 3 x – 3
(ii) Express as a single fraction in its simplest form:
3
2
_____
+ _____
x–2 2–x
30
Worke d E x a m p l e 2. 10
3a2 – 6a
_______
2–a
=
=
1
3a(a – 2)
_________
–1 (a – 2)
3a
___
–1
= –3a
1
2
Solution
3x – 1
3x – 1
2
4
2
4
+ _____ = _____ – ___________ + _____
(i) _____ – __________
x–3
x – 1 x2 – 4x + 3 x – 3 x – 1 (x – 1) (x – 3)
(2)(x – 3) – (3x – 1)(1) + 4(x – 1)
= ___________________________
(x – 1)(x – 3)
2x – 6 – 3x + 1 + 4x – 4
= _____________________
(x – 1)(x – 3)
3x – 9
= ___________
(x – 1)(x – 3)
3(x – 3) 1
= ___________
(x – 1)(x – 3) 1
3
_____
=
x–1
3
2
(ii) _____ + _____
x–2 2–x
3
2
–1
= _____ + _____ ⋅ ___
x – 2 2 – x –1
3
2
= _____ – _____
x–2 x–2
3–2
= _____
x–2
1
_____
=
x–2
ALGEBRA I
Exercise 2.3
Express as single fractions in their simplest form:
x+y
4x – 1 2x – 5
10. _______
1. ______ + ______
4
2
3x + 3y
3x – 7 ______
5x – 3
______
ab3
–
2.
11. ___
4
12
a2b
2x – 4
x – 1 ______
_____
x+y
+
3.
5
3
12. _______
x2 – y2
7
______
+2
4.
ab + b2
4x – 5
13. _______
a2 – b2
2
1
5. ___ – ___
p–q
5x 7x
14. _____
q–p
8
6. _____ + x
x+1
x2 – 2x
15. ______
3
1
_____
__
x–2
–
7.
x+5 x
15x2 + 5x
16. ______________
5
2
_____
______
+
8.
15x2 + 20x + 5
x – 2 3x – 1
x3 – 8
3
1
17. ___________
2
9. ______ + ______
x + 2x + 4
3x + 5 2x – 1
3
4
1
25
25. _____ – _____ + ________
x + 2 x – 3 x2 – x – 6
3
1
4
26. ___________
+ _____ – _____
x2 + 3x + 2 x + 2 x + 1
3
a + 1 _____
a–1
27. _____ – ______
+
a – 1 a2 – 1 a + 1
a
a
28. _____ + _____
a+b a–b
a–3
a+2
29. _____ + _____
a–2
a+3
p+1 p+2
30. ______ + ______
p–2
p–1
x3 – 8
18. ___________
3
x + 2x + 4
x3 + y3
19. _______
x2 – y2
x4 – y4
20. ______
x2 – y2
(p – q)3
21. _______
p2 – q2
16x
x2 – 8x + 16 ____
–
22. ___________
x2 – 16
16x2
x
2x
23. _____ – _____
x–1 1–x
x–5
x+4
24. ______
+ ______
x2 –16 x2 – 25
1
1 ______
1
__
31.
31 ______
3
n+1+n+n+2
a + b ______
a–b
32.
3 ______
– 2
2
2
a – b b – a2
a–b
a+b
33. ____________
3
– _____________
a2 – 2ab + b2 a2 + 2ab + b2
a
,
34.
3 Write as a single fraction in the form _____
b
x –c
where a, b and c ∈ N:
5
1
(i) _____
+ _____
x2 – 1 1 – x2
3x
18
– ___________
(ii) ___________
x2 + 3x – 18 18 – 3x – x2
31
2
2 .4 ALGEBR AIC FR ACTIONS II:
2.4
MULTIPLICATION AND DIVISION
Complex Fractions
Sometimes a fraction’s numerator and/or denominator may be a fraction as well. These types of fractions
are called complex fractions.
ALGEBRA I
1 + _13
For example: _____
3
_
_
2
5+7
Wo rke d Exa m p le 2 . 1 2
Worke d E x a m p l e 2. 13
4x – 3
3x2 – 27
× _______.
Simplify ____________
2
4x – 15x + 9 x2 + 3x
x2 – 25
x–5
.
Simplify _____ ÷ ___________
2
x + 1 x + 4x + 3
Solution
Solution
3x2 – 27
_____________
4x2 – 15x + 9
×
4x – 3
_______
x2 – 25
÷ ___________
x + 1 x2 + 4x + 3
x2 + 4x + 3
x–5
= _____ × ___________
x+1
x2 – 25
x–5
_____
x2 + 3x
3(x2 – 9)
(4x – 3)
= ____________ £ ________
(4x – 3)(x – 3) x (x + 3)
1
=
1
1
1
3(x +3) (x – 3) (4x – 3)
____________________
=
(4x – 3)(x – 3)(x)(x + 3)
1
1
1
3
= __
x
(x – 5)
______
(x + 1)
£
1
=
1
(x + 3)(x + 1)
____________
(x + 5)(x – 5)
1
x+3
_____
x+5
Wo rke d Exa m p le 2 . 1 4
9
1 – __2
x
Simplify ______ .
6
2 + __
x
Solution
9
x2 – 9
Numerator: 1 – __2 = _____
x
x2
(x – 3)(x + 3)
= ____________
x2
Alternative Method
A
2x + 6
6 ______
Denominator: 2 + __
x=
x
2(x + 3)
= _______
x
9
1 – __2
2(x + 3)
(x – 3)(x + 3) _______
x
______
∴
÷
= ____________
x
6
__
x2
2+ x
1
=
(x – 3)(x + 3)
____________
x–3
= _____
2x
32
x2
Multiply the numerator and the denominator by x2.
M
9
1 – __2
x
x2
______
=
· __2
6 x
2 + __
1 x
x2 – 9
= ________
2x2 + 6x
1
1
=
1
×
x
_______
2(x + 3)
1
=
(x + 3) (x – 3)
____________
2x (x + 3)
x–3
_____
2x
1
2
Exercise 2.4
For Questions 1 to 19, express as single fractions in their simplest form.
25a3
10b2 _____
1. _____
×
2b
5a2
4
2
3. ______ ÷ ______
2x – 1 4x2 – 1
4x – 4 _____
x2 – 1
5. ______
÷
5
x
x3
2x2 – x – 1 ______
4x2 – 1
7. __________
∙
2x2 + x – 1 x2 – 1
4x 8x3
2. ___ ÷ ___2
2y 4y
2x + 6
x2 + x – 2
4. __________
× ______
4x – 4
x2 + 2x – 3
2y2 – 8y
y2 – 64 _______
×
6
6. _______
2y – 16
y2 – 16
3x
8x2 – 34x – 9
8. ____________ × ________
4x + 1
4x2 – 81
2
9x2 – 16x
6x2 – 20x + 16 ____________
÷ 2
9. _____________
2
4x – 16x + 16 2x + 2x – 12
2–
x2
________
x
x2 – 6x + 9
__________
15. _________
5
x–7
(3 – x)(x – 3)
x–y
____
xy
_____
11. __
5
a – _1b
16. ______
1 – _1b
x–5
____
x+5
_______
12. _____
1
p+q
_____
1
______
17. __1 _1
___
x
+1
x+1
_______
13.
3
___
y–x
18. ______
2
xy
p
x2 – 25
+
19.
2
x –y
______
x2
_____________
2
2
x + 2xy + y
____________
x
1
1
2 __
20. If u = x – __
x and v = x – x2,
show that u2(u2 + 4) = v2.
a3 – b3
a3 – ab2
21. If x = ________
and y = ________
,
3
2
a + ab
a2b + b3
x b(a2 + ab + b2)
.
show that __y = ______________
a2(a + b)
ALGEBRA I
4x + 3
_____
x2 – 49
________
10.
16x2 – 9
______
_____
1+1
x
14. ______
__
1
q
x+3
____
x + 3x
_____
x2 – y2
x+1
2 .5 LONG DIVISION IN ALGEBR A
2.5
Another approach to dividing one algebraic expression by another is to use long division.
Worke d E
Exa
x a m p le
le 2.15
Simplify:
6x3 + 5x2 – 10x + 3
__________________
2x –1
Solution
3x2 + 4x – 3
2x – 1
6x3
–
(6x3
+
–
3x2)
We multiply 2x –1 by 3x2 to get 6x3 –3x2 and subtract to get 8x2.
8x2 – 10x
Bring down the next term. We divide 8x2 by 2x to get 4x.
– 10x + 3
– (8x2 – 4x)
We multiply 2x –1 by 4x to get 8x2 –4x and subtract to get –6x.
–6x + 3
– (–6x + 3)
0
∴
Arrange the divisor and dividend in descending powers of x.
We divide 6x3 by 2x to get 3x2.
5x2
6x3 + 5x2 – 10x + 3
__________________
2x – 1
Bring down the next term. We divide –6x by 2x to get –3.
We multiply 2x –1 by –3 to get –6x + 3 and subtract.
We get a remainder of 0.
= 3x2 + 4x – 3
33
2
Wo rke d Exa m p le 2 . 1 6
Divide x3 – 4x2 + 3 by x – 1.
Solution
As the x term is missing, we write 0x into the expression.
x2 – 3x – 3
x – 1 x3 – 4x2 + 0x + 3
– (x3 – x2)
–3x2 + 0x
ALGEBRA I
– (–3x2 + 3x)
–3x + 3
– (–3x + 3)
0
∴ (x3 – 4x2 + 3) ÷ (x – 1) = x2 – 3x – 3
Long division is also useful as it can be used to find the other factors of an expression.
Consider the following question: ‘Find the prime factors of 30.’
Starting with the smallest prime factor of 30, which is 2, we have:
30 ÷ 2 = 15
Continuing:
15 ÷ 3 = 5
So the prime factors of 15 are 3 and 5
∴ 30 = 2(3)(5). Its prime factors are 2, 3 and 5.
Long division in algeba is very similar to finding the prime factors of a positive whole number like 30.
Wo rke d Exa m p le 2 . 1 7
Show that x – 4 is a factor of 12x3 – 55x2 + 29x – 4 and find the other two factors.
Solution
12x2 – 7x
There is no remainder, so (x – 4) is a factor.
T
+1
x – 4 12x3 – 55x2 + 29x – 4
= (x – 4) (4x – 1)(3x – 1)
– (12x3 – 48x2)
∴ The other two factors are (4x – 1) and (3x – 1).
– 7x2 + 29x
– (–7x2 + 28x)
x–4
– (x – 4)
0
34
∴ 12x3 – 55x2 + 29x – 4 = (x – 4) (12x2 – 7x + 1)
2
Exercise 2.5
9. Divide x5 – x4 – 6x3 – 8x2 + 8x + 48 by
x2 – x – 6. Hence, fully factorise
x5 – x4 – 6x3 – 8x2 + 8x + 48.
Simplify each of the following:
1. (x3 + 2x2 – 7x – 2) ÷ (x – 2)
2. (3x3 + 13x2 – 18x – 40) ÷ (3x + 4)
10.
1 Show that x – 1 is a factor of
2x4 – 14x3 + 22x2 – 10x.
3. (6x3 – 29x2 + 21x – 4) ÷ (3x – 1)
1
11. Find all four factors of 36x4 – 289x2 – 400,
if 4x2 – 25 divides evenly into the quartic
expression.
4. (14x3 + 33x2 – 5x) ÷ (7x – 1)
5. (36x3 – 28x + 8) ÷ (3x – 2)
6. (15x4 – 11x3 – 77x2 + 31x + 42) ÷ (5x + 3)
8. Find the quotient when:
4x4 – 8x3 – 13x2 + 2x + 3 is divided by 2x – 1.
Hence, find the other factors.
13.
1 Find all four factors of
72x4 + 18x3 – 29x2 – 3x + 2, if 6x2 + x – 2
divides evenly into the quartic expression.
1 Investigate if g(x) = 3x + 4 and h(x) = 5x + 4
14.
are factors of the function
f(x) = –6x4 – 29x3 – 16x2 + 16x.
ALGEBRA I
7. Divide x3 + 17x + 18 by x + 1.
Hence, find the other factors.
12.
1 Show that 2x + 1 is not a factor of
6x3 – 13x2 – 19x + 12.
Give a reason for your answer.
Revision Exercises
1. (a) Factorise the following:
(iii) (4x – 5)3
(i) x2 – x – 90
(iv) (4p + 3)3 (p – 2)
(ii) 4x2 + 4x + 1
(v) 2a(3a + 4)2(2a + 5)3
(iii) 10x2 – x – 2
(iv) 9x2 – 12xy + 4y2
(v) 14x2 – 15x + 4
(b) Simplify:
3a – 9b
(i) ________
6a – 18b
8x – 10
(ii) _________
16x2 – 25
a3 – b3
(iii) _______
6a – 6b
by + b – y – 1
(iv) _____________
b3 – 1
x2 – x
(v) __________
2
x – 4x + 3
(c) Expand:
(i) (3x + 7)2
(
)
1
1 _____
(d) Simplify __
x – x + h ÷ h.
2. (a) Factorise the following:
(i) x2 – 100
(ii) 4x2 – 81
(iii) 25x2 – 49y2
(iv) 121a2 – 144b2
(v) (5x + 6y)2 – (x + y)2
(b) Write as a single fraction in
its simplest form:
5
3
(i) _____ + _____
x–3 x+4
6
7
(ii) ______ – ______
2y + 1 2y – 1
2
x
(iii) _____ – _____
x+1 x–1
(c) Using Pascal’s triangle,
expand the following
polynomials:
(i) (2a + 1)5
(ii) (4b – 7c)4
(iii) (6x – 5)6
2(9 – 2x)
2
(d) Show that _____ + ________
x–4
4–x
simplifies to a constant for
all x ∈ R, x ≠ 4.
3. (a) Factorise:
(i) x3 + 27
(ii) a3 + 8
(iii) b3 + 1000
(iv) 8x3 + 125
(v) 125x3 + 27y3
(ii) (2x – 1)3
35
2
(b) Simplify the following:
2x – 2y
(iv) _______
3y – 3x
2x2 + 9x + 4
(i) _____________
2x2 + 11x + 5
a2 – b2
(ii) ______
a3 – b3
a2 – b2
(iii) ______
a4 – b4
x–3
(v) _____2
9–x
3x
5x2 + 9
2x
(c) Write _____ + _____ – _______
x+3 x–3
x2 – 9
k
as a fraction in the form _____
x+t
where k, t ∈ Z.
6. (a) Simplify each of the following:
(i) (4x3 – 5x2 – 2x + 3) ÷ (x – 1)
(ii) (36x3 – 18x2 – 10x + 4) ÷ (3x – 1)
(c) Write as single fractions in their lowest
terms:
ALGEBRA I
5
1
(i) _____ + _____
x–2 2–x
5
7
(ii) ______ + ______
2y – 1 1 – 2y
(iii) (24x4 – 22x3 – 7x2 + 4x + 1) ÷ (3x + 1)
(iv) (2x4 + 13x3 + 19x2 – 10x – 24)
÷ (2x + 3)
2
2
(iii) _____ + _____
b–a a–b
9
4
(iv) ______ – ______
2x – 1 1 – 2x
4. (a) Factorise:
(i) y3 – 1
(iv) x3 – 216
(ii) 8y3 – 1
(v) 1000a3 – 343b3
(iv) (36x5 + 117x4 + 95x3 + 5x2 – 11x – 2)
÷ (3x – 1)
(b) (i) Show that 8x + 1 is a factor of
48x3 + 62x2 – 33x – 5 and find the
other factors.
(iii) 27a3 – 8b3
(b) Express as single fractions in their lowest
terms:
6x
1
– _____
(i) _____
x2 – 9 x + 3
2y
_______
4
(ii) ______ – 2
3y – 2 9y – 4
(c) The sides of a triangle are (m2 + n2),
(m2 – n2) and 2mn in length.
(i) Prove that the triangle is right-angled.
(ii) Show that 4x – 2 is not a factor of
8x3 + 22x2 – 7x – 3.
(c) Write down the dividend, divisor and
quotient when
(6x5 – 40x4 – 26x3 + 356x2 – 312x – 144)
is divided by (3x2 + 10x + 3).
7. (a) Rearrange the order of the terms in these
expressions and then factorise them:
(i) 51 + x2 + 20x
(ii) –x2 + 169
m2 + n 2
(iii) ax + by + ay + bx
2 mn
(iv) a2 + b2 – 2ab
(v) a2 + b2 – c2 – 2ab
m2 – n2
(ii) Deduce the three lengths of the sides
if m = 5 and n = 2.
x
(b) (i) Write 1 + __y as a single fraction.
(
x
(ii) Hence, simplify 1 + __y
5. (a) Factorise fully:
(i) 3x2 – 75
(ii) 9x3 – 25x
(iv) x4 – 81
(v) ax2 – bx2 – ay2 + by2
(iii) x4 – y4
(b) Simplify:
ax + ay – cx – cy
(i) ________________
ax + ay + cx + cy
9y3
–y
(ii) ___________
3
3y + 8y – 3
(a + b)2 – c2
(iii) ___________
a2 – (b + c)2
36
(c)
)(
2
y
______
y2
– x2
)
.
(i) Show that x – 1 is not a factor
of 4x4 – 20x3 – 7x2 + 32x + 15.
Give a reason for your answer.
(ii) Show that 4x2 – 1 is a factor of
12x4 + 4x3 – 59x2 – x + 14 and find
the other factors.
(iii) Show that x3 – x2 is a factor of
4x5 – 7x3 + 3x2 and find the other
factors.
8. (a) Factorise fully the following expressions:
(d)
1
y – __
y
(i) x3 + y3 + 3x + 3y
(ii) x2 – y2 + 5x + 5y
10. (a)
1
(v) a2 – (b + c)2
x2 – 2xy + y2 – z2
9
2x – 4
× ______
(i) _______________
(iii) ________
27
x2 – y2 – 2yz – z2
4x2 – 16
x2 – 8x
______
x–5
(c) Car A leaves a town and drives north at a
speed of 40 km/hr. Car B leaves 3 hours later
and travels north along the same route at a
speed of 60 km/hr. After a certain distance,
Car B catches up with Car A. Write an
expression for this distance for each car.
5
8
(d) Write _______ – _______
6x – 15 4x – 10
a
in the form ________ where a, b ∈ Z.
b(2x – 5)
9. (a) Factorise the following:
y2
x2
(i) ____ – ___
100 49
1
(ii) x3 + __3
x
(iv)
8y2
– 64
÷
3t – 10
8
11. (a) (i) Show that _______ – ______
1
3t – 2 2 – 3t
2
simplifies to a constant, t ≠ __.
3
3
8x + 27
(ii) Simplify ________
.
4x2 – 9
x2 – x – 6
(iii) Simplify ________
.
x2 – 4
(ii) a2 + b2 – c2 + 2ab
(b) Simplify:
(i)
5b2
10ab
and y = ______
(b) If x = ______
,
a2 – ab
b2 – a2
x –b(b + a)
show that __y = _________
.
2a2
z
z
_____
+ _____
z–1 z+1
______________
(c) Simplify _____
z
z
– _____
z–1 z+1
(i) a2 + 2ab + b2
(c) Show that
(v) x2 – y2 + 2x + 1
15x2
____
(iii) Hence, factorise
x3 + y3 + z3 + 3xy(x + y).
(b) Factorise:
(iii) x4 – 16
a2b2
(i) Factorise x3 + y3.
ALGEBRA I
×
)
(ii) Hence, write x3 + y3 + 3xy (x + y) as a
perfect cube (i.e. as (expression)3).
(b) Simplify the following:
x2 – 64
2
2 + _____
y–1
( )(
(iv) x3 – y3 + x2 – y2
(ii)
and
1
2
_____
(ii) Simplify y – __
y 2+y–1.
(iii) x2 – 2xy + y2 – 4z2
x2 – 25
______
2
(i) Write as single fractions:
5x3
___
2
y2
yz
+ ___________ + ___________
(x – y) (x – z) (y – z) (y – x) (z – x) (z – y)
x
simplifies to _____.
(x – z)
x2
___________
1
__
(2x2 – 50)
9
____________
(ii)
1
___
2
45
(4x – 20x)
(c) The area of a rectangle can be expressed as
2x2 + 3x – 20. The length of the rectangle
is x + 4.
(i) Find the breadth of the rectangle in
terms of x.
This rectangle is used as a base for
a rectangular box. The volume of
this box can be expressed as
6x3 + 7x2 – 63x + 20.
1
1
1 (a) Write _____ – ______ as a single fraction
12.
x + 2 2x – 1
and hence solve:
1
1
1
_____
– ______ = ___________
x + 2 2x – 1 2x2 + 3x – 2
k
(b) Express in the form _____:
x+3
5
20
4
_____
+ _____ – _________
x – 2 x + 3 x2 + x – 6
(c) Factorise 8x3 – 27y3 + 4x2 – 9y2.
(ii) Find the height of this rectangular box in
terms of x.
37