F2 1st Term Maths Teaching Material (1st Edition)

S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny

Unit 1 Estimation and Approximation

A. IntroductionMeasurements only give approximate values. Instruments with a finer

graduation give a more precise result. Precision have to be specified if we

want to get the same result in measurements.

B. Measurement with different precisionsExample 1.1

Measure the length of the line segments.

A B

X Y

Precisions

Line segments 0.1cm 0.5cm cm

AB 8.8cm 9.0cm 9cm

XY 5.4cm 5.5cm 5cm

C. Numerical Estimation1. Reformulation

Rounding

Round off the numbers to the nearest unit before doing the calculation.

Example 1.2

3.775+2.145-2.603

≈4+2-3

=3

1

Difference between “9.0cm” and “9cm”As an estimated value, the precision of “9.0cm” is less than 1cm and that of “9cm” is or greater than 1cm. In other words, “9.0cm” is more precise.


Compensation

Choose a clustered value to represent all the numbers for estimation.

Example 1.3

61+58+63+59+57+59

≈60+60+60+60+60+60

=60x7

=420

Using Compatible numbers

Change the numbers to make the estimation easier.

Example 1.4

0.3356x181

≈(1 3)x180

=60

2. Compensation

Make adjustments in calculations.

Example 1.5

2217x8.03

=(2000x8.03)+(217x8.03)

≈(2000x8)+(200x8)

=16000+1600

=17600

Note: The remains 217x8.03 is also considered so that the estimated value

will be more precise.

3. Translation

Change the structure of the problem.

Example 1.6

(56.4x3) 22

=56.4x(3 22)

≈56x(3 21)

=56 7

=8

2


4. Rounding up/ Rounding Down

Rounding up

Choose slightly greater values to replace the original values.

Example 1.7a

13+26+18+22+34+48+15

≈20+30+20+30+40+50+20

=210

Rounding down

Choose slightly smaller values to replace the original values.

Example 1.7b

13+26+18+22+34+48+15

≈10+20+10+20+30+40+10

=140

Note: Rounding up is to estimate the greatest possible value of the sum of

numbers and rounding down is to estimate the smallest possible value of it.

Therefore, it is very important to know whether the greatest or smallest

possible value should be estimated before answering some long questions.

Example 1.8

There are 8 people in a lift. The maximum loading of the lift is 600kg and the

people’s weights are 67.3kg, 88.5kg, 53.2kg, 41.1kg, 107.3kg, 56.6kg, 48.9kg

and 50.3kg respectively. Estimate whether the lift will be overload.

Solution:

Rounding up the weights,

The sum of weight

=67.3+88.5+53.2+41.1+107.3+ 56.6+48.9+50.3

≈70+90+60+50+110+60+50+60

=550kg

Since the estimated value is lower than 600kg, the lift will not be overload.

Note: As all the weights of the people have been rounded up, the estimated

sum of weight must be greater than the real value of the sum. Therefore, if the

estimated sum is less than the maximum loading, the lift will not be overload.

3


Example 1.9

Peter, John, Joe and James have $7.1, $6.5, $8.4 and $9.9 respectively. A

football costs $30. Do they have enough money to buy it?

Solution:

Rounding down the values,

The sum of money

=7.1+6.5+8.4+10.9

≈7+6+8+10

=31

Since the estimated sum of money is greater than the price of the football,

they have enough money to buy it.

Note: Rounding down the amount of money each person has, the real sum of

money is greater the estimated one. As the estimated sum is greater than the

price of football, they must have enough money to buy it.

D. Significant numbersOne method of rounding off a number is to take their nearest units. Another

method is to take a few significant numbers (or significant figures) from

the number and delete all the less important digits.

Rules of selecting significant figures from a number

Numbers Rule

For numbers greater than 1 Starting from the first digit, all the digits in the

number are significant.

For numbers between 0 and 1 Digits starting from the first non-zero digit are

significant.

Example 1.10a

Find the first significant figure of the following numbers:

2.737154 1504331 0.093234 1.05113 0.90911

Solution:

2.737154 1504331 0.093234 1.05113 0.90911

4


Example 1.10b

Find the first three significant figures of the following numbers:

2.737154 1504331 0.093234 1.05113 0.90911

Solution:

2.737154 1504331 0.093234 1.05113 0.90911

Rounding off numbers to a certain number of significant figures

To round off a number, we can take the first few significant figures from a

number.

Example 1.11

Correct the following numbers to four significant figures.

637547 1.630299 0.00903993 1.800012 17.7371

Solution:

637500 1.630 0.009040 1.800 17.74

E. ErrorsAbsolute Error

The absolute error is the difference between the actual value and the

measured value. The error is “absolute” because it is always positive.

Example 1.12

Find the absolute errors of the following cases.

The length of a string The weight of a person

Actual value 13.75cm 115.3kg

Measured value 14cm 115kg

Solution:

The length of a string The weight of a person

Actual value 13.75cm 115.3kg

Measured value 14cm 115kg

Absolute error 14cm-13.75cm

=0.25cm

115.3kg-115kg

=0.3kg

5

Upper Limit = Measured value + (Max. absolute error)Lower Limit = Measured Value - (Max. absolute error)


Maximum absolute error, upper and lower limit

However, since the actual figures are usually unknown, we often have to find

out the maximum absolute error so that we can obtain the range of values

of the actual value.

When the precision of a measurement and the measured value are stated, we

can find out the upper limit and lower limit of the actual figure, as well as the

maximum absolute error. The greater the maximum absolute error, the less

precise the figure is.

For example, the length of a line is estimated to be 9cm, correct to the nearest

cm.

In the range of 8.5cm and 9.5cm, the estimated length of the line will be 9cm.

Therefore, the actual length of the line will between 8.5cm and 9.5cm because

only such a range of values can be estimated at 9cm. The upper and lower

limits of the actual value will be 9.5cm and 8.5cm respectively.

6

Maximum Absolute Error = ( Upper limit – Lower limit ) 2


Example 1.13

Find the upper limit, the lower limit and the maximum absolute error of the

actual values.

Height of a person Weight of a ball

Measured value 173.0cm 793g

Correct to the nearest… 0.5cm 1g

Solution:

Height of a person Weight of a ball

Measured value 173.0cm 793g

Correct to the nearest… 0.5cm 1g

Upper Limit 173.25cm 793.5g

Lower Limit 172.75cm 792.5g

Maximum absolute error (173.25-172.75) 2

=0.5cm

(793.5-792.5) 2

=1g

Relative error & percentage error

A relative error and percentage error can represent the comparison between

the absolute error and the true value or that between the maximum

absolute error and the measured value.

Note: The value of a relative error and the corresponding percentage error are

the same. However, a relative error is represented by a fraction or a decimal

number and a percentage error is represented by a percentage.

7

Relative Error = or

Percentage error = Relative error 100%


Example 1.14a

The weight of a watermelon is measured to be 2.73kg, and the actual weight

of it is 2.7255kg. Find

i) The absolute error

ii) The relative error (corr. to 3 sig. fig.)

iii) The percentage error (corr. to 3 sig. fig.)

Solution:

i) The absolute error

=2.73-2.7255

=0.0045kg

ii) The relative error

=

=0.001651073…

=0.00165 (corr. to 3 sig. fig.)

iii) The percentage error

= x100%

=0.1651073....%

=0.00165% (corr. to 3 sig. fig.)

Example 1.14b

The weight of a watermelon is measured to be 2.73kg, correct to the nearest

0.01kg. Find

i) The upper and lower limit

ii) The maximum absolute error

iii) The relative error (corr. to 3 sig. fig.)

iv) The percentage error (corr. to 3 sig. fig.)

i) The upper limit

=2.735kg

The lower limit

=2.725kg

8


ii) The maximum absolute error

=(2.735-2.725) 2

=0.005kg

iii) The relative error

=

=0.00183150183…

=0.00183 (corr. to 3 sig. fig.)

iv) The percentage error

= x100%

=1.83150183…%

=1.83% (corr. to 3 sig. fig.)

F. AbbreviationsCorrect to 3 significant figures corr. to 3 sig. fig.

Correct to 2 decimal places corr. to 2 d.p.

G. Exercise1. Round off the following numbers.

corr. to 3 d.p. corr. to 3 sig.

fig.

corr. to 4 sig.

fig.

corr. to the

nearest

tenth

1.25703

0.89099

12.7356

0.004504504

corr. to the

nearest 10

corr. to the

nearest 1000

corr. to 3 sig. fig.

49509

290999

195034

13507

2. Round off 2090.999 to

9


the nearest ten

the nearest integer

1 decimal place

2 significant figures

3. Estimate the following.

i) 9.9+10.4+10.6+9.5+10.1+10.2+9.7

ii)

iii)

iv)

v)

10


vi)

vii)

viii)

ix)

x)

11


xi)

4. Complete the following table.

Time for running 100m Length of a chopstick

Absolute error

Measured value 12.93s 9.81cm

Actual value 12.937s 9.83cm

Relative error (in a form

of fraction)

Percentage error (corr.

to 3 sig. fig.)

5. The length of a piece of string is estimated at 10.3cm, correct to the

nearest 0.1cm.

i) Find the range of the actual length.

ii) Find the maximum absolute error.

12


iii) Find the relative error (corr. to 3 sig. fig. when necessary).

iv) Find the percentage error (corr. to 3 sig. fig. when necessary).

6. The length, width and height of a rectangular board are 12.7cm, 15.8cm

& 18.3cm respectively, correct to the nearest 0.1cm.

i) Find the greatest possible volume of the board.

ii) Find the least possible volume of the board.

13


iii) Find the maximum absolute error of the volume of the board.

iv) Find the relative error and the percentage error of the volume of the

board (corr. to 3 sig. fig.).

v) Find the maximum absolute error of the total surface area of the

board.

14


<<END OF UNIT 1>>

Unit 2 Polynomials

A. MonomialsA monomial is a polynomial containing 1 term only and satisfies one of the

following conditions.

Conditions Examples

It contains numbers only.-8, 0, 45, 2.7,

It is a product of number and

variables. The power of the variable

must be positive integers

, -6xy, 3x

Examples of non-monomials

Conditions Examples

The expression contains more than

1 term.

,

The expression is divided by a

variable, i.e. the denominator of the

expression contains a variable.

, ,

The powers are variables. ,

Degree of monomials

Conditions Method to find out the

degree

Example Degree of

monomial

Containing

numbers only

The degree of the

monomial is 0 because

0

15


the power of the

variables is 0.

Product of

numbers and one

variable

The degree of the

monomial equals to the

power of the variable.

3

4 (as

)

Product of

numbers and

more than one

variable

The degree of the

monomial equals to the

sum of power of the

variables.

1+5=6

1+5+2=8

B. PolynomialsPolynomials are the sum of 1 or more than 1 monomial.

Monomial, binomial and trinomial

Polynomials Definition Examples

Monomial Polynomial containing 1

term only,

Binomial Polynomial containing 2

terms,

Trinomial Polynomial containing 3

terms

,

Terms of polynomials

Polynomials can be separated into different terms. It can be done by finding

all the monomials in the polynomial.

Examples Terms No. of terms

2x, +y, +1 3

, 2

, , -3y, 4

1

Degree of polynomials

The degree of a polynomial equals to the degree of the term with the highest

degree.

16


Example 2.1

Find the degree of the polynomial .

Solution:

There are four terms in the polynomial. They are , , and -3.

Terms Degree

1+2=3

1+1=2

5

-3 0

Since has the highest degree and its degree is 5, the degree of the

polynomial is also 5.

Coefficients of a term

The coefficient of a term is the number multiplied to the variables. The term

with no variables is called a constant term.

Example 2.2

Find the coefficients of the terms in the polynomial.

Polynomials Coefficient Constant

term

17


Solution:

Polynomials Coefficient Constant

term

1 0 -4 0 0

0 0 4 0 0 7

C. Addition and subtraction of polynomialsThe addition and subtraction of polynomials can be done by adding up the

coefficients of the like terms together.

Like terms

Like terms are the terms having the same variables. Their coefficients can be

different. Here are some examples of like terms and unlike terms.

Like terms Reasons

, Same variable multiplied.

,

,

Unlike terms Reasons

, Even though the coefficients are the

same, they are not like terms if they

have different variables multiplied.

, Same variables but with different

powers cannot be regarded as like

terms.,

Addition and subtraction of polynomials in a row

This can be done by adding or subtracting the like terms in the polynomials. If

there are brackets, they have to be removed first.

18


Example 2.3

Simplify the following.

Solution:

=

=

Example 2.4

Simplify the following.

Solution:

=

=

Note: Since and are not like terms, they cannot be added together.

Addition and subtraction of polynomials in a column

This can be done by putting the like terms in the same column, and the

addition or subtraction can be done.

Example 2.5

Evaluate the following.

Solution:

19


Example 2.6

Evaluate the following.

Solution:

D. Multiplication of polynomialsDistributive Law

Some multiplication of polynomials can be done with the distributive law.

Formula of multiplication of polynomials

Consider the figure above. Find the total area of the rectangle.

First solution:

Area of the figure

=Area of I + Area of II + Area of III + Area of IV=

Second solution:

Area of the figure

20

The distribution law

I

III

II

IV

a

b

c d


=Width x Length

=

From the solutions above, since the area of the rectangle doesn’t change, we

can conclude that .

Also from the distributive law, considering (a+b) as one term,

=(a+b)c+(a+b)d

=

Conclusion: formulas for multiplying polynomials

They are:

1.

2.

Example 2.7

Expand .

Solution:

=

Example 2.8

Expand .

Solution:

Using the distributive law,

=

=

=

Or using the formula

21

With the distributive law,

3x

2

21x

2


=

=

=

Example 2.9

Expand .

Solution:

=

=

=

=

=

Example 2.10

Expand .

Solution:

=

=

=

=

=

Expand the first two brackets at

the beginning.

Treat the first two terms in each

bracket as one term.

E. Exercise1. Complete the following table.

Polynomials No. of Coefficients Constant

22


terms termx xy y

2. Classify the following expressions.

4a 4x-3-ay -9 4ab2+bc-d a3+b 2 m a10+b+c7 b2-4ac

(a) Monomials :

(b) Binomials :

(c) Trinomials :

(d) None of the above:

In question 3 to 5, arrange the terms of the following polynomials in

descending powers of the variables.

3. 3y + 2 – 5y3 + y5 =

4. 7 + 3x3 – x – x5 =

5. 4c2 – 5c3 + 3 + 2c =

In question 6 to 8, arrange the terms of the following polynomials in ascending

powers of the variables.

6. p4 + 3p3 – 5p2 + p =

7. 2x3 – 4x6 + 7 – 9x =

8. –r2 + 8 – 10r5 – 6r3 =

Expand and simplify the following.

9.

10.

23


11.

12.

13.

14.

24


15.

16.

17.

18.

19.

25


20.Complete the following table. Put a tick under “monomial”, “binomial”,

“trinomial” and write down the degree of polynomial and number of terms if

they are polynomials. If not, put a tick under the column “not a polynomial”.

Algebraic

expressions

Polynomial Not a

polynomial

Degree of

polynomial

Number

of

terms

Monomial Binomial Trinomial

881903

<<END OF UNIT 2>>

Unit 3 Use of Formulae

A. Algebraic fractionAlgebraic fraction is a fraction in which the denominator is not a constant, i.e.

contains variables. Also, it is assumed that the denominator in an algebraic

fraction is not equal to 0.

, and are some examples of algebraic fractions.

B. Addition and subtraction of algebraic fractionsThe addition and subtraction of algebraic fractions can be done by the

following steps.

For example, now we have to simplify the algebraic fraction .

Ste

p

Description Example

1 Find the lowest common multiple of

the denominators. The lowest common

multiple should be obtained by

26


multiplying the fraction with a non-zero

number.=

=

2 Add or subtract the numerators.

=

3 Simplify the fraction.

=

Example 3.1

Simplify .

Solution:

=

=

=

Example 3.2

Simplify .

Solution:

Simply add the numerators when the

denominators are the same.

Simplify the sum of the fractions.

27


=

=

Example 3.3

Simplify .

Solution:

=

=

=

Example 3.4

Simplify .

Solution:

=

=

=

Example 3.5

Simplify .

Solution:

The lowest common multiple of and

is . Multiply by 2 to

obtain .

28


=

=

=

=

=

If there is not any common factor between two

denominators, the common denominator will be

the product of the two denominators.

C. Basic concept of factorizationThe idea of factorization is to take out the common factor from different terms.

The common factor may be a number or a variable.

Factorization by taking common factors

A number or a monomial is multiplied by some factors. Common factors

indicate the same numbers or variables contained in both monomials at the

same time.

For example, in the algebraic expression ,

The factors of pq are p and q. (1 is also a factor of this monomial, but it is

meaningless to be taken out in factorization.)

The factors of qr are q and r.

Therefore they have a common factor q.

29


Taking out the common factor,

=

Therefore the result of factorizing is .

Example 3.6Factorize .

Solution:

=


Solution:

=

=


Solution:

=

=

Example 3.9

Factorize .

Solution:

2 is the common of the factor of 2a and 2b.

2 is a factor of 6 because .

30


=

=

=

=

Example 3.10

Factorize .

Solution:

=

=

=

Example 3.11

Factorize .

Solution:

=

=

=

Treat (n-1) as a term.

In the expression , and

.You may convert the expression into

this form to avoid mistakes.

Take out the common factors of the variables.

Take out the common factors of the numbers.

D. Formulae and substitutionSubstitution is to put the values of the variables into a formula.

Example 3.12

Given that , find the value of S

when a=3, b=7 and n=-2.

Solution:

=

=

Example 3.13

Given that , find the value of b

Simply substitute the values

into the values to find the

unknown.

31


when a=2, c=-7 and d=-1.

Solution:

E. Change the subject of the formula

Do you still remember the formula of distance - , or ?

This formula helps us to find the speed when the distance and time are given.

However, when the value we want to find is the distance but not the speed,

we have to change the subject of the formula to help us find it out more

easily.

Example 3.14

Change the subject of

to a.

Solution:

Example 3.15

Change the subject of the

following formula.

Put the terms containing the subject to the left

side in order to separate them from the other

terms.

Divide the both side by the coefficient (-3) of

the subject (a).

Usually, the denominator of a fraction will not

be a negative number.

The symbol [w] means to change the subject

32


Solution:

Example 3.16

Change the subject of the

following formula.

Solution:

of the formula to w.

Take out the terms containing the subject (w).

Put the terms containing the subject to the left.

Divide both sides by the term multiplied by the

subject.

When there are fractions in the formula,

multiply both side by the lowest common

multiple of the denominator.

Put the terms containing the subject to the left.

Take out the subject from the terms.

Divide both sides by the term multiplied by the

subject.

F. Exercise1. Factorize the following expressions.

a. b.

33


c. d.

e. f.

g. h.

2. Simplify the following expressions.

a. b.

c. d.

34


e. f.

g. h.

3. Change the subject of the following formulae.

a. b.

c. d.

35


e. f.

g. h.

4. The Body Mass Index (BMI) is used to indicate whether a person is

obese or not. The relationship between the BMI (B), the weight of a

person in kilograms (m) and his height in meters (h) is .

a. Change the subject of the formula to m.

b. Given that the BMI of a healthy person is 21. If his height is 1.7m, using

the formula in (a), find his ideal weight.

36


c. Amy’s height is 160cm, find her weight in terms of B.

d. Given that people with their BMI over 23 is regarded is overweight. If

Mary is 155cm tall and weighs 120 pounds, is she overweight (given that

1kg is equal to 2.2 pounds)?

e. Alex’s height is 210cm, if his weight is 90kg, find the difference between

the BMI of Alex and Mary (corr. to 3 sig. fig.)

f. Hiromi is overweight. Her height is 160cm and her weight is 70kg. She

wants to decrease her BMI to 21.5. Find the weight she needs to lose in

order to achieve this target.

37


5. It is known that the volume of a cone can be calculated with the formula

, where V, r and h represents the volume (in ), radius of the

base (in cm) and the height (in cm) respectively.

a. When , and , find V.

b. When the base radius and the volume are 3cm and 66 respectively,

find the height of the cone. (Take )

<<END OF UNIT 3>>

Unit 4 Identities and factorization

A. The concept of identityAn equation is only true for one or a few special values of the unknowns.

However, an identity is true for all values of the unknowns.

How to prove and disprove an identity

An identity can be proved or disproved with the following steps.

Example 4.1

Determine whether the equation is an

identity.

38


Solution:

L.H.S. =

=

=

=

R.H.S =

Therefore, L.H.S.=R.H.S.

Therefore the equation is an identity.

Example 4.2

Determine whether the equation is an

identity.

Solution:

L.H.S. =

=

=

R.H.S.=

=

Therefore, L.H.S R.H.S.

Therefore the equation is not an

identity.

Expand both sides to see

whether the coefficient of every

term is the same.

Finding a counter example is also a method to disprove an identity. However,

since an identity satisfies all values of unknowns, giving examples is not

enough to prove an identity. Therefore, giving an example can only disprove

an identity but cannot prove one.

Example 4.3

Determine whether the following

equation is an identity.

Solution:

When ,

L.H.S. =

Substitute one value of the

unknown into the both sides of the

39


=

R.H.S.

=4

Since L.H.S R.H.S,

the equation is not an identity.

equation. If the value of both sides

are the same, use the method in

example 4.1 and 4.2 to prove or

disprove the identity and do not

use this method to prove an

identity. You may continue to try

other values if you think that the

equation is not an identity.

Comparing coefficients of like terms

You may use this method to find some unknowns in the identity.

Example 4.4

If , find the

values of A and B.

L.H.S. =

=

=

=

Therefore, = .

Therefore, A=5 and B=6.

Expand the left hand side so the

coefficients of the terms of both

sides can be compared.

Method of substitution

This method can also find some unknowns in an identity. Since it is an identity,

the coefficients of all terms are always unchanged. Therefore this method can

be used.

Example 4.5

If , find the

values of A and B.

Solution:

When x=0,

L.H.S. =

40


=5

R.H.S.

=B

Therefore, B=5.

Therefore the identity is

.

When ,

L.H.S. =

=

R.H.S.

=

Since both L.H.S.=R.H.S.,

Therefore A=6.

A=6, B=5

Find a value so that one of the

unknown will be multiplied by 0 (so

that it will be eliminated).

”1” is used because this makes

the calculation easier. You may also

substitute other values.

B. Some important algebraic identitiesThe following identities can be used to expand and factorize some algebraic

expressions.

1.

[Proof]

L.H.S. =

=

=

=

=

R.H.S.

41


,

is true.

2.

[Proof]

L.H.S. =

=

=

=

=

=

R.H.S.

is true.

3.

[Proof]

L.H.S. =

=

=

=

=

=

R.H.S.

is true.

Expanding expressions using the algebraic identities

Example 4.6

Expand .

Solution:

=

=

Example 4.7

Expand .

Use the identity .

Treat and .

42


Solution:

=

=

Example 4.8

Expand .

Solution:

=

=

Example 4.9

Expand .

Solution:

=

=

Example 4.10

Expand .

Solution:

=

=

=

=

=

Use the identity .

Treat and . Add a bracket

outside 4p and 3q in order to show a

clear work and avoid mistakes.

Use the identity .

Treat and .

Use the identity

. Treat and .

Use the identity .

Treat and .

Use the identity .

Treat and .

43


C. FactorizationFactorization, the reverse of expansion, is the process rewriting an algebraic

expression in a form of the product of its factors. An algebraic expression

can be factorized by taking common factors, grouping terms, using

algebraic identities. The method of taking common factors has been

introduced in Unit 3 so only the remaining two will be talked about in this unit.

Factorization by grouping terms

Not every term in an algebraic expression has one common factor. There may

be two groups of terms that have their own common factors.

Example 4.11

Factorize .

Solution I:

=

=

=

Solution II:

=

=

=

Example 4.12

Factorize .

Solution:

=

=

=

Example 4.13

Factorize .

Put pq and qr in one group, ap and ar

into another group.

Taking out the common factors from both

groups.

Taking out the common factor -

from both groups.

You may also put pq and ap into one

group, qr and ar into another group.

Separate the terms into 2 groups.

Take out the common factors from the

groups.

Take out the common factor .

44


Solution:

=

=

=

=

Example 4.14

Factorize .

Solution:

=

=

=

=

=

Example 4.15 (Non-essential)

Factorize

.

Solution:

=

=

=

=

Arrange the terms to make the

calculation easier.

Find the common factors of all terms

first, and then separate the terms into

groups.

Take out the common factor among

these six terms.

Group the terms into groups. (Other

groupings are also possible)

45


Example 4.16

Factorize .

Solution:

<Trial I>

=

=

<Trial II>

=

=

Therefore, this expression cannot be

factorized.

Take out the common factor

.

Try to separate the first and last two

terms into two groups first.

However, after taking the common

factors from each bracket, there are no

common terms left.

Try to put the first and third term into one

group and the remaining ones into another.

However, there is still no common factor

left. Therefore the expression cannot be

factorized.

Note: Before using the method of grouping terms, it is necessary to check

whether there are any common factors of all terms.

Factorization by using identities

We have learnt three identities. They are:

1. or

2. or

3. or

We can apply these identities to factorize algebraic expressions apart from

expanding them.

Example 4.17

Factorize .

46


Solution:

=

=

Example 4.18

Factorize .

Solution:

=

=

Example 4.19

Factorize .

Solution:

=

=

=

=

Example 4.20

Factorize .

Solution:

=

=

Example 4.21

Factorize .

Solution:

Change the terms in the form of .

Use the identity . Treat

and .



and .

Take out the common factor of the terms before

using the identities.



and .

Change the expression in the form of .


and .

47


=

=

=

Example 4.22

Factorize .

Solution:

=

=

Example 4.23

Factorize .

Solution:

=

=

=

=

Example 4.23 (Non-essential)

Factorize .

Solution:

=

=

=

=

=

=

Take out the common factor of all terms. Take -1

out of the expression so that the coefficient of

won’t be a negative number.



and .



and .



and .



and .

Factorize the first three terms first.

Change first three terms in the form of

.


and .

Change expression in the form of .


and .

48


D. Simplification, multiplication and division of algebraic fractionsThe simple addition and subtraction of algebraic expression has been

introduced in the previous unit, and we will go further in this unit.

Simplification of algebraic fractions

The simplification of algebraic fractions is to cancel the common factors

between the numerator and the denominator.

Example 4.24

Simplify .

Solution:

=

=

=

Example 4.25

Simply .

Solution:

=

=

Take out the common factor mn from the

numerator and the denominator.

Cancel the common factor.

Take out the common of the terms of the

numerator.

Cancel the common factors between the

49


=

=

Example 4.26

Simplify .

Solution:

=

=

=

=

Example 4.27

Simplify .

Solution:

=

=


Factorize the numerator and the

denominator.

Cancel the common factor from

the numerator and the denominator.

Cancel the common factor b from the



denominator by taking out the common

factors in all terms.


50


=

=

=

denominator using identities.

Cancel the common factors.

Multiplication and division of algebraic fractions

With factorization, we can simplify the product and quotient of algebraic

fractions and make the calculation easier.

Example 4.28

Simplify .

Solution:

=

=

=

Example 4.29

Simplify .

Solution:

=

=

Cancel the common factors between

the numerator and denominator of the

fractions.

Change all the division signs to times

signs first.

Factorize the numerators and

denominators of all fractions.

51


=

=

=

Example 4.30

Simplify .

Solution:

=

=

=

=

=

=

Cancel the common factors.

Change the division sign to a times

sign first.

Factorize the numerators and

denominators of the fractions.

Cancel all the common factors.

E. Exercise1. Expand the following.

a. b.

52


c. d.

e. f.

g. h.

i. j.

53


2. Determine whether the following equations are identities.

a.

b.

c.

3. Factorize the following.

a. b.

54


c. d.

e. f.

g. h.

i. j.

k. l.

55


m. n.

o. p.

q. r.

s. t.

4. If , find A and B.

56


5. If , find A and B.

6. Answer the following questions.

a. Factorize .

b. Factorize .

c. Using the result of (a) and (b), factorize .

7. Simplify the following.

a. b.

57


c. d.

e. f.

g. h.

58


i. j.

k. l.

<<END OF UNIT 4>>

Unit 5 Pythagoras’ Theorem

A. Square and square rootsThere are two rules about square roots:

59


1. If , then x is the square root of y.

2. The square roots of are and . is the positive root and is

the negative root.

Example 5.1

Find the square roots of 16, 0.64,

and .

Solution:

and , the

square roots of 16 are 4 and -4.

and ,

the square roots of 0.64 are 0.8

and -0.8.

,

,

The square roots of are

and .

is a negative number,

there is no real square roots of

.

There are no real square roots for a

negative number.

B. Values of square rootsSome square roots cannot be expressed as a fraction or an integer when its

square is not a perfect square number. They can only be expressed as surd

forms. The numbers that can only be expressed as surd forms are surds.

60


Examples of surd forms: , and .

Estimation of square roots

We can use two consecutive square numbers to estimate the value of a

square root. If , then .

Example 5.2

Find two consecutive numbers

between which lies.

,

Therefore lies between 1 and 2.

Example 5.3

Find two consecutive numbers

between which lies.

,

Therefore lies between 14 and

15.

Find two consecutive perfect square

numbers between which 3 lies.

Operations of square roots

There are two rules about the operations of square roots

1. , where and .

2. , where and .

Example 5.4

Simplify .

Solution:

61

,,,


=

=

=

Example 5.5

Simplify .

Solution:

=

=

=

Example 5.6

Simplify .

Solution:

=

=

=

=

Example 5.7

Simplify .

Solution:

Express in the form of

product of prime factors.

62


=

=

=

=

=2

C. Pythagoras’ TheoremPythagoras’ Theorem

In , if , then . (Pyth. Theorem)

Converse of Pythagoras’ Theorem

In , if , then . (Converse of Pyth. Theorem)

Example 5.8

Determine whether the following

triangles are right-angled triangles.

63


Solution:

In ,

=

=

=

=

=

,

is a right-angled triangle.

(Converse of Pyth. Theorem)

In ,

=

=

=

=

=144

, is not a

right-angled triangle.

Example 5.9

Only check whether the sum of

squares of the shorter sides

is equal to the square of

the longest side . Other

groupings such as and

are not needed.

As , only check

whether is correct

or not.

64


Find AC.

Solution:

In ,

(Pyth. Theorem)

Example 5.10

Find the unknown.

Solution:

In ,

(Pyth. Theorem)

Example 5.11

Even though is also

correct, as AC must be positive, only

is possible.

65


A ladder is leaning against the wall.

The distance between the foot of

the ladder and the top of the wall is

4m and the height of the wall is

2.4m. Find the distance between

the foot of the ladder and the wall.

Solution:

Let the distance be m.

(Pyth. Theorem)

Therefore the distance is 3.2m.

Remember to add back the unit

after solving the equation.

D. Rationalization of denominators

66


We can rationalize the denominator of a fraction so that its denominator will

not be an irrational number.

Example 5.12

Rationalize .

Solution:

=

=

Example 5.13

Rationalize .

Solution:

=

=

=

=

=

Multiply the denominator by .

Check if the denominator is a

rational number after the fraction is

rationalized.

Multiply the denominator by .

E. Exercise

67


1. Simplify the following. Rationalize the denominator if necessary. Your

answer should be in a surd form.

a. b.

c. d.

e. f.

g. h.

68


2. Find two consecutive numbers between which the following numbers.

a.

b.

3. Rationalize the denominator of the following fractions.

a. b.

c. d.

69


e. f.

4. Determine whether the following triangles are right-angled triangles.

a.

b.

70


c.

5. Find the unknowns in the figures.

a.

b.

c.

71


d.

6. ABCDEFGH is a cuboid.

a. Find BD.

b. Find BE.

72


7. There is a ladder lying against the wall. The length of the ladder is 4.2m

and its foot is 2.5cm apart from the wall. If it moves 0.5m farther from the

wall, how far will the top of the ladder slide down? (correct your answer to

3 significant figures)

8. Mr. Wong bought a TV. Its length is 65cm and its width is 58cm, correct

to the nearest cm. Find the upper and lower limit of the length of its

diagonal. (correct your answer to 3 significant figures)

73


<<END OF UNIT 5>>

Unit 6 Analysis of Statistical Graphs

A. Grouping of dataFrequency distribution

Suppose the scores of a group of 50 students has been recorded and shown

in the following table.

25 71 44 58 61 73 90 81 61 57

40 45 42 62 74 68 77 70 49 51

53 60 77 48 33 39 88 22 49 47

56 55 60 62 83 79 78 54 51 49

48 51 53 90 82 60 52 48 30 42

A few steps should be done in order to see the trend of the data.

1. Find the largest and smallest values of the data.

From the data above, the largest value and the smallest value are 90 and 22

respectively.

2. Decide the number of classes and the range of each class. The

classes should cover all data.

Suppose the range of each class is 10. They are 21-30, 31-40, 41-50, 51-60,

61-70, 71-80 and 81-90. You may also use other ranges but the classes

should cover all data.

3. Use tallies to record the number of data in each class.

The following table shows the data separated into different classes.

Score Tally Frequency

21-30 3

31-40 3

41-50 11

51-60 14

74


61-70 6

71-80 7

81-90 6

B. Terminologies about classesThere are some terminologies about classes.

(1) Range – The difference between the maximum and the minimum value of

the data. The largest value and the smallest value are 90 and 22

respectively. Therefore the range in this example is (90-22)=68.

(2) Class interval – Intervals with equal length cover the range of data

between the maximum and the minimum without overlapping. For

example, the class interval of the first class is 21-30.

(3) Class limit – Two ends of each class interval. The smaller value is the

lower class limit and the greater value is the upper class limit. For

example, the lower class limit of the second class is 31 and the upper

class limit is 40.

(4) Class mark – The average of the upper class limit and the lower class

limit. For example, the class limit of the second class is .

(5) Lower class boundary – The minimum possible value in the class. For

example, the lower class boundary of the second class is 30.5.

(6) Upper class boundary – The maximum possible value in the class. For

example, the upper class boundary of the second class is 40.5.

(7) Class width – The difference between the upper class boundary and the

lower class boundary. For example, the class width of the second class is

.

75


C. HistogramThe data in a histogram is presented by rectangular bars with no gaps and

each bar is divided by its class boundaries. The following figure is the

histogram showing the score of the 50 students referring to the table in part A.

Before drawing the histogram, we use this frequency distribution table to

present the data.

Score Class mark Frequency

21-30 25.5 3

31-40 35.5 3

41-50 45.5 11

51-60 55.5 14

61-70 65.5 6

71-80 75.5 7

81-90 85.5 6

Something that you should be careful with:

(1) A title must be given.

(2) The vertical axis (y-axis) represents frequency.

(3) The horizontal (x-axis) and vertical axes (y-axis) should be labelled clearly

76


with suitable units.

(4) There should be no gaps between 2 bars. Each bar is perpendicular to

the horizontal axis (x-axis).

(5) Only class mark is marked for each class.

(6) The area of each bar is proportional to the frequency of the

corresponding class. If the class width of all classed are equal, the height

of the bar is proportional to the frequency of the corresponding class.

D. Frequency polygons and frequency curvesWe can also use a frequency polygon to present the grouped data. Two class

intervals with zero frequency have to be added to both ends of the data when

constructing the frequency polygon. Therefore, the frequency distribution

polygon based on the scores of the 50 students will be as follows:

Process of constructing a frequency polygon:

1. Construct a frequency distribution table with the class mark of each class.

2. Mark the class mark of each class on the x-axis. Mark the frequency on

the y-axis.

3. Give the title to the frequency polygon.

4. Mark the frequency corresponding to its class mark on the graph.

5. Join the points to form a frequency polygon.

A frequency polygon can also be drawn by smoothing the frequency polygon.

77


Something you should be careful with:

1. If a frequency polygon is marked with the crosses (X), the frequencies can

be read from them and the total number of data can be calculated. If not,

we can only know the trend of the data but cannot know the frequencies.

Therefore, in examinations, when constructing a frequency polygon, the

crosses should be marked unless special instructions are given.

2. We can construct two or a few frequency polygons in one graph in order to

compare the sets of data but we cannot use a histogram to do so.

E. Cumulative frequency polygons and cumulative frequency curvesA cumulative frequency polygon or cumulative frequency curve is used to

present cumulative quantities. The following table shows the height of 50

students.

Height (cm) Class boundaries (cm) Frequency

141-145 140.5-145.5 3

146-150 145.5-150.5 4

151-155 150.5-155.5 4

156-160 155.5-160.5 6

161-165 160.5-165.5 12

166-170 165.5-170.5 11

171-175 170.5-175.5 7

176-180 175.5-180.5 3

78


The frequency distribution table above can be changed to the following

cumulative frequency table:

Height less than (cm) Cumulative frequency

140.5 0

145.5 3

150.5 7

155.5 11

160.5 17

165.5 29

170.5 40

175.5 47

180.5 50

From the table, we can know the number of students who get less than a

certain score. Note that the phrase “cumulative frequency” is used to

describe the number of data with their values lower than a certain level.

Using the data above, we can construct a cumulative frequency polygon.

(Note that the unit should be added to the frequency distribution table and the

label of the x-axis.)

Also, a cumulative frequency curve can be obtained.

79


F. PercentilesThe percentile indicates the percentage of data with their values lower than

that datum. The following graph is the cumulative frequency polygon of the

height of 100 students.

For example, when we want to find the weight for which 40% of the students’

weights are under it (i.e. to find the 40th percentile, , we have to do the

following steps:

1. Find its corresponding cumulative frequency.

The cumulative frequency

=

=40

2. Find the weight corresponding to the cumulative frequency.

80


From the yellow line, the weight is 55.5kg. Therefore . Do you

know how to find out with the graph above?

Also, some percentiles can be represented by quartiles. They are

a. The lower quartile, , where .

b. The median, , where .

c. The upper quartile, , where .

Example 6.1

The cumulative frequency polygon shows the weight of 50 students. Find

(a) ;

(b) the median;

(c) .

Solution:

81


(a)


=

=12.5

From the graph above,

.

(b)


=

=25

From the graph above, the

median is 58kg.

(c)


=

=45

From the graph above,

.

G. Abuse of statisticsThese are the common abuses of statistics:

1. The scales of the axes are adjusted to exaggerate the relative change.

Description Example

The values of

the markings

on the y-axis

do not start

from 0.

2. The sizes of the figures are overcastted to exaggerate the differences.

Description Example

82


A bar in the

chart is wider

than the others

so that the area

of the bar will

be much larger

than the others.

3. There may be hidden information in the figures which would lead the

readers to make wrong conclusions.

Description Example

Only a part of

the values are

compared. (The

sales of

magazines with

higher rankings

are not shown.)

83


Regard some

values as

“others”. (In this

pie chart, some

of the “other”

candidates may

have more

supporters than

Candidate B

does.)

Only present the

”percentage

increase” but not

the actual value.

(The actual

sales of the soft

drinks are not

shown.)

Example 6.2

The following chart shows the percentage increase of sales of different brands

of video games.

84


a. List two misleading points of the bar chart.

b. State the objective of the bar chart.

a. The vertical axis does not start from zero. Therefore it overstates the

percentage increase of sales Ouendan. Also, only the percentage

increase, but not the actual sales, is shown, we cannot conclude which

video game sells better.

b. To let the readers think that Ouendan sells better.

H. Exercise1. Here are the weights, in kg, of 50 newborn babies in King’s Hospital.

2.3 1.9 3.4 3.5 3.2 2.9 3.6 2.8 1.8 3.1

3.0 3.1 3.3 2.4 2.7 1.7 1.8 2.4 0.9 1.6

85


2.3 2.2 3.0 3.2 3.1 3.0 4.1 3.7 3.3 1.4

3.5 3.6 3.7 2.7 2.9 2.5 2.5 2.3 2.0 2.1

2.4 2.9 3.4 4.2 3.3 2.8 2.6 1.7 4.0 2.2

a. Use the above data to complete the following table.

Weight

(kg)

Class boundaries

(kg)

Class mark

(kg)

Tally Frequency

0.6-1.0

1.1-1.5

1.6-2.0

2.1-2.5

2.6-3.0

3.1-3.5

3.6-4.0

4.1-4.5

b. Construct a frequency polygon using the above data.

c. Construct a histogram using the above data.

d. Construct a cumulative frequency polygon using the above data.

e. Find the first quartile, median and using the cumulative frequency

polygon.

2. The following frequency polygon shows the monthly income, in $1000, of

100 people. Given that the class interval of the first class is $0-$5000.

86


a. Which class interval has the highest frequency?

b. How many people have the income between $20000.5 and $24999.5?

c. Find the number of people who has the income below $14999.5.

d. Use the above data to complete the following table.

Income lower than ($) Cumulative frequency

5000

10000

15000

20000

25000

30000

35000

e. Use the above data to construct a cumulative frequency polygon and find

the median.

87


3. The following graph shows the sales of different magazines.

a. From the figure above, what can you say about the sales between 88

Magazine and the others?

b. State two misleading points of the graph.

4. Here is the cumulative frequency polygon showing the scores of F.2A and

F.2B students in a mathematics test.

a. Find the lower quartile, median and upper quartile of the score of F.2A

and F.2B students respectively.

88


b. What conclusion can you make between the result of F.2A and F.2B

students?

<<END OF UNIT 6>>

89

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F2 1st Term Maths Teaching Material (1st Edition)