Contents

M: : 1 ALGEBRAIC EXPRESSIONS
1.1.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.2 THE REAL NUMBER SYSTEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.3 RATIONAL AND IRRATIONAL NUMBERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.4 ROUNDING OFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.1.5 ESTIMATING SURDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.1.6 PRODUCTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.1.7 FACTORISATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.1.8 SIMPLIFICATION OF FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.1.9 CHAPTER SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1
 M : 1 Algebraic expressions

1.1.1 Introduction EMA2

Over human history, all peoples and cultures have contributed to the field of Mathematics. Topics like algebra
may seem obvious now, but for many centuries mathematicians had to make do without it. Over the next three
grades, you will explore more advanced and abstract mathematics. It may not be obvious how this applies to
everyday life, but the truth is, mathematics is required for nearly everything you will do one day. Enjoy your
mathematical journey. Remember, there is no such thing as a “maths person” or “not a maths person”. We can
all do mathematics, it just takes practice.

Figure 1.1: Some examples of early tally sticks. These were used to help people count things such as the number of days
between events or the number of livestock they had.

In this chapter, we will begin by revising the real number system and then learn about estimating surds and
rounding real numbers. We will also be expanding on prior knowledge of factorisation and delve into more
complex calculations involving binomial and trinomial expressions.

1.1.2 The real number system EMA3

Real R

Rational Q
Integer Z
Whole N0 Irrational Q’

Natural N

We use the following definitions:

• N: natural numbers are {1; 2; 3; . . .}

• N0 : whole numbers are {0; 1; 2; 3; . . .}
• Z: integers are {. . . ; −3; −2; −1; 0; 1; 2; 3; . . .}

2 1.1.1 Introduction
3
VISIT:
The following video shows an example of determining which of the above sets of numbers a particular number
is in.
See video: 2DBH at www.everythingmaths.co.za

NOTE:
Not all numbers are real
√ numbers.
√ The√square root of a negative number is called a non-real or imaginary
number. For example −1, −28 and −5 are all non-real numbers.

1.1.3 Rational and irrational numbers EMA4

DEFINITION: Rational number

A rational number (Q) is any number which can be written as:

a
b
where a and b are integers and b ̸= 0.

The following numbers are all rational numbers:

10 21 −1 10 −3
; ; ; ;
1 7 −3 20 6

We see that all numerators and all denominators are integers.

This means that all integers are rational numbers, because they can be written with a denominator of 1.
DEFINITION: Irrational numbers

Irrational numbers (Q′ ) are numbers that cannot be written as a fraction with the numerator and denominator
as integers.

Examples of irrational numbers: √

√ √ √3 1+ 5
2; 3; 4; π;
2

These are not rational numbers, because either the numerator or the denominator is not an integer.

Decimal numbers EMA5

All integers and fractions with integer numerators and non-zero integer denominators are rational numbers.
Remember that when the denominator of a fraction is zero then the fraction is undefined.

You can write any rational number as a decimal number but not all decimal numbers are rational numbers.
These types of decimal numbers are rational numbers:
4
• Decimal numbers that end (or terminate). For example, the fraction 10 can be written as 0,4.
• Decimal numbers that have a repeating single digit. For example, the fraction 13 can be written as 0,3̇ or
0,3. The dot and bar notations are equivalent and both represent recurring 3’s, i.e. 0,3̇ = 0,3 = 0,333....
2
• Decimal numbers that have a recurring pattern of multiple digits. For example, the fraction 11 can also
be written as 0,18. The bar represents a recurring pattern of 1’s and 8’s, i.e. 0,18 = 0,181818....
NOTE:
You may see a full stop instead of a comma used to indicate a decimal number. So the number 0,4 can also
be written as 0.4

M : 1. Algebraic expressions 3
4
Notation: You can use a dot or a bar over the repeated digits to indicate that the decimal is a recurring decimal.
If the bar covers more than one digit, then all numbers beneath the bar are recurring.

If you are asked to identify whether a number is rational or irrational, first write the number in decimal form.
If the number terminates then it is rational. If it goes on forever, then look for a repeated pattern of digits. If
there is no repeated pattern, then the number is irrational.

When you write irrational numbers in decimal form, you may continue writing them for many, many decimal
places. However, this is not convenient and it is often necessary to round off.

NOTE:
Rounding off an irrational number makes the number a rational number that approximates the irrational num-
ber.

Worked example 1: Rational and irrational numbers

QUESTION

Which of the following are not rational numbers?

1. π = 3,14159265358979323846264338327950288419716939937510...
2. 1,4
3. 1,618033989...
4. 100
5. 1,7373737373...
6. 0,02

SOLUTION

1. Irrational, decimal does not terminate and has no repeated pattern.

2. Rational, decimal terminates.
3. Irrational, decimal does not terminate and has no repeated pattern.
4. Rational, all integers are rational.
5. Rational, decimal has repeated pattern.
6. Rational, decimal has repeated pattern.

Converting terminating decimals into rational numbers EMA6

A decimal number has an integer part and a fractional part. For example, 10,589 has an integer part of 10 and
a fractional part of 0,589 because 10 + 0,589 = 10,589.

Each digit after the decimal point is a fraction with a denominator in increasing powers of 10.

For example:
1
• 0,1 is 10
1
• 0,01 is 100
1
• 0,001 is 1000

4 1.1.3 Rational and irrational numbers

5
This means that
5 8 9
10,589 = 10 + + +
10 100 1000
10 000 500 80 9
= + + +
1000 1000 1000 1000
10 589
=
1000

VISIT:
The following two videos explain how to convert decimals into rational numbers.
Part 1
See video: 2DBJ at www.everythingmaths.co.za
Part 2
See video: 2DBK at www.everythingmaths.co.za

Converting recurring decimals into rational numbers EMA7

When the decimal is a recurring decimal, a bit more work is needed to write the fractional part of the decimal
number as a fraction.

Worked example 2: Converting decimal numbers to fractions

QUESTION

a
Write 0,3̇ in the form b (where a and b are integers).

SOLUTION

Step 1: Define an equation

Let x = 0,33333...

Step 2: Multiply by 10 on both sides

10x = 3,33333...

Step 3: Subtract the first equation from the second equation

9x = 3

Step 4: Simplify

3 1
x= =
9 3

M : 1. Algebraic expressions 5
6
Worked example 3: Converting decimal numbers to fractions

QUESTION

Write 5,4̇3̇2̇ as a rational fraction.

SOLUTION

Step 1: Define an equation

x = 5,432432432...

Step 2: Multiply by 1000 on both sides

1000x = 5432,432432432...

Step 3: Subtract the first equation from the second equation

999x = 5427

Step 4: Simplify

5427 201 16
x= = =5
999 37 37

In the first example, the decimal was multiplied by 10 and in the second example, the decimal was multiplied
by 1000. This is because there was only one digit recurring (i.e. 3) in the first example, while there were three
digits recurring (i.e. 432) in the second example.

In general, if you have one digit recurring, then multiply by 10. If you have two digits recurring, then multiply
by 100. If you have three digits recurring, then multiply by 1000 and so on.
√
Not all decimal numbers can be written as rational numbers. Why? Irrational decimal numbers like 2 =
1,4142135... cannot be written with an integer numerator and denominator, because they do not have a pattern
of recurring digits and they do not terminate.

Exercise 1 – 1:

1. The figure here shows the Venn diagram for the special sets N, N0 and Z.
X

Z
N0

a) Where does the number − 12

3 belong in the diagram?

6 1.1.3 Rational and irrational numbers

7
 b) In the following list, there are two false statements and one true statement. Which of the statements
is true?
i. Every integer is a natural number.
ii. Every natural number is a whole number.
iii. There are no decimals in the whole numbers.
2. The figure here shows the Venn diagram for the special sets N, N0 and Z.

Z
N0

a) Where does the number − 12 belong in the diagram?

b) In the following list, there are two false statements and one true statement. Which of the statements
is true?
i. Every integer is a natural number.
ii. Every whole number is an integer.
iii. There are no decimals in the whole numbers.
3. State whether the following numbers are real, non-real or undefined.
√
√ 0 √ − 7 √ √
a) − 3 b) √ c) −9 d) e) − −16 f) 2
2 0
4. State whether the following numbers are rational or irrational. If the number is rational, state whether it
is a natural number, whole number or an integer.
√
9
a) − 13 b) 0,651268962154862... c)
3
√
d) π 2 e) π 4 f) 3 19
(√
3
)7
g) 1 h) π + 3 i) π + 0,858408346

5. If a is an integer, b is an integer and c is irrational, which of the following are rational numbers?
5 a −2 1
a) b) c) d)
6 3 b c
a
6. For each of the following values of a state whether is rational or irrational.
14
√
a) 1 b) −10 c) 2 d) 2,1
7. Consider the following list of numbers:
√ 4 √ 22 14 √ 7
−3 ; 0 ; −1 ; −8 ; − 8; ; ; 7 ; 1,34 ; 3,3231089... ; 3 + 2 ; 9 ; π ; 11
5 7 0 10

Which of the numbers are:

a) natural numbers b) irrational numbers
c) non-real numbers d) rational numbers
e) integers f) undefined

M:1 Algebraic expressions 7

8
 8. For each of the following numbers:
• write the next three digits and
• state whether the number is rational or irrational.
a) 1,15̇ b) 2,121314... c) 1,242244246...
d) 3,324354... e) 3,32435̇4̇
9. Write the following as fractions:
a) 0,1 b) 0,12 c) 0,58 d) 0,2589
10. Write the following using the recurring decimal notation:
a) 0,1111111... b) 0,1212121212... c) 0,123123123123... d) 0,11414541454145...
11. Write the following in decimal form, using the recurring decimal notation:
25 10 7 2
a) b) c) d)
45 18 33 3
3 5 1
e) 1 f) 4 g) 2
11 6 9
12. Write the following decimals in fractional form:
a) 0,5̇ b) 0,63̇ c) 0,4̇ d) 5,31 e) 4,93 f) 3,93

For more exercises, visit www.everythingmaths.co.za and click on ’Practise Maths’.

1. 2DBM 2. 2DBN 3a. 2DBP 3b. 2DBQ 3c. 2DBR 3d. 2DBS 3e. 2DBT
3f. 2DBV 4a. 2DBX 4b. 2DBY 4c. 2DC2 4d. 2DC3 4e. 2DC4 4f. 2DC5
4g. 2DC6 4h. 2DBZ 4i. 2DBW 5. 2DC7 6. 2DC8 7. 2DC9 8a. 2DCB
8b. 2DCC 8c. 2DCD 8d. 2DCF 8e. 2DCG 9a. 2DCH 9b. 2DCJ 9c. 2DCK
9d. 2DCM 10a. 2DCN 10b. 2DCP 10c. 2DCQ 10d. 2DCR 11a. 2DCS 11b. 2DCT
11c. 2DCV 11d. 2DCW 11e. 2DCX 11f. 2DCY 11g. 2DCZ 12a. 2DD2 12b. 2DD3
12c. 2DD4 12d. 2DD5 12e. 2DD6 12f. 2DD7

1.1.4 Rounding off EMA8

Rounding off a decimal number to a given number of decimal places is the quickest way to approximate a
number. For example, if you wanted to round off 2,6525272 to three decimal places, you would:

• count three places after the decimal and place a | between the third and fourth numbers;
• round up the third digit if the fourth digit is greater than or equal to 5;
• leave the third digit unchanged if the fourth digit is less than 5;
• if the third digit is 9 and needs to be rounded up, then the 9 becomes a 0 and the second digit is rounded
up.

So, since the first digit after the | is a 5, we must round up the digit in the third decimal place to a 3 and the
final answer of 2,6525272 rounded to three decimal places is 2,653.

8 1.1.4 Rounding off

9
Worked example 4: Rounding off

QUESTION

Round off the following numbers to the indicated number of decimal places:

120
1. = 1,1̇2̇ to 3 decimal places.
99
2. π = 3,141592653... to 4 decimal places.
√
3. 3 = 1,7320508... to 4 decimal places.
4. 2,78974526 to 3 decimal places.

SOLUTION

Step 1: Mark off the required number of decimal places

If the number is not a decimal you first need to write the number as a decimal.

120
1. = 1,212|121212 . . .
99
2. π = 3,1415|92653 . . .
√
3. 3 = 1,7320|508 . . .
4. 2,789|74526

Step 2: Check the next digit to see if you must round up or round down
1. The last digit of 120
99 = 1,212|1212121̇2̇ must be rounded down.
2. The last digit of π = 3,1415|92653 . . . must be rounded up.
√
3. The last digit of 3 = 1,7320|508 . . . must be rounded up.
4. The last digit of 2,789|74526 must be rounded up.
Since this is a 9 we replace it with a 0 and round up the second last digit.

Step 3: Write the final answer

120
1. = 1,212 rounded to 3 decimal places.
99
2. π = 3,1416 rounded to 4 decimal places.
√
3. 3 = 1,7321 rounded to 4 decimal places.
4. 2,790

Exercise 1 – 2:

1. Round off the following to 3 decimal places:

a) 12,56637061... b) 3,31662479... c) 0,2666666...
d) 1,912931183... e) 6,32455532... f) 0,05555555...
2. Round off each of the following to the indicated number of decimal places:
a) 345,04399906 to 4 decimal places.
b) 1361,72980445 to 2 decimal places.
c) 728,00905239 to 6 decimal places.

M:1 Algebraic expressions 9

10
 1
d) to 4 decimal places.
27
45
e) to 5 decimal places.
99
1
f) to 2 decimal places.
12
3. Study the diagram below

A π B

E D π C

a) Calculate the area of ABDE to 2 decimal places.

b) Calculate the area of BCD to 2 decimal places.
c) Using you answers in (a) and (b) calculate the area of ABCDE.
d) Without rounding off, what is the area of ABCDE?
r
4. Given i = ; r = 7,4; n = 96; P = 200 000.
600
a) Calculate i correct to 2 decimal places.
b) Using you answer from (a), calculate A in A = P (1 + i)n .
c) Calculate A without rounding off your answer in (a), compare this answer with your answer in (b).
5. If it takes 1 person to carry 3 boxes, how many people are needed to carry 31 boxes?
6. If 7 tickets cost R 35,20, how much does one ticket cost?

For more exercises, visit www.everythingmaths.co.za and click on ’Practise Maths’.

1a. 2DD9 1b. 2DDB 1c. 2DDC 1d. 2DDD 1e. 2DDF 1f. 2DDG 2a. 2DDH 2b. 2DDJ
2c. 2DDK 2d. 2DDM 2e. 2DDN 2f. 2DDP 3. 2DDQ 4. 2DDR 5. 2DDS 6. 2DDT

1.1.5 Estimating surds EMA9

√
If√the nth root of a √
number cannot be simplified to a rational number, we call it a surd. For example, 2 and
3
6 are surds, but 4 is not a surd because it can be simplified to the rational number 2.
√ √ √
a is any positive number, for example,√ 7 or 3 5.
In this chapter we will look at surds of the form n a where √
It is very common for n to be 2, so we usually do not write 2 a. Instead we write the surd as just a

It is sometimes useful to know the approximate value √ of a surd without having to use a calculator. For example,
we
√ want to be able to estimate where a surd like
√ 3 is on the number line. From a calculator we know that
√ 3 is equal to 1,73205.... It is easy to see that 3 is above 1 and below 2. But to see this for other surds like
18, without using a calculator you must first understand the following:

√ √
n
If a and b are positive whole numbers, and a < b, then n
a< b

10 1.1.5 Estimating surds

11
A perfect square is the number obtained when an integer is squared. For example, 9 is a perfect square since
32 = 9.

Similarly, a perfect cube is a number which is the cube of an integer. For example, 27 is a perfect cube, because
33 = 27.
√
3
√ √
Consider the surd 52. It lies somewhere between 3 and 4, because 3 27 = 3 and 3 64 = 4 and 52 is between
27 and 64.

Worked example 5: Estimating surds

QUESTION
√
Find two consecutive integers such that 26 lies between them. (Remember that consecutive integers are two
integers that follow one another on the number line, for example, 5 and 6 or 8 and 9.)

SOLUTION

Step 1: Use perfect squares to estimate the lower integer

√
52 = 25. Therefore 5 < 26.

Step 2: Use perfect squares to estimate the upper integer

√
62 = 36. Therefore 26 < 6.

Step 3: Write the final answer

√
5 < 26 < 6

Worked example 6: Estimating surds

QUESTION
√
3
Find two consecutive integers such that 49 lies between them.

SOLUTION

Step 1: Use perfect cubes to estimate the lower integer

√
33 = 27, therefore 3 < 3 49.

Step 2: Use perfect cubes to estimate the upper integer

√
43 = 64, therefore 3 49 < 4.

Step 3: Write the answer

√
3 < 3 49 < 4

Step 4: Check the answer by cubing all terms in the inequality and then simplify
√
27 < 49 < 64. This is true, so 3 49 lies between 3 and 4.

M:1 Algebraic expressions 11

12
Exercise 1 – 3:

1. Determine between which two consecutive integers the following numbers lie, without using a calculator:
√ √ √
3
√ √
a) √ 18 b) √ 29 c) √ 5 d) √3 79 e) √ 155
3 3
f) 57 g) 71 h) 123 i) 90 j) 3 81
2. Estimate the following surds to the nearest 1 decimal place, without using a calculator.
√ √ √ √
a) 10 b) 82 c) 15 d) 90
3. Consider the following list of numbers:
√ √ √ √
27
7 ; 19 ; 2π ; 0,45 ; 0,45 ; − 4 ; 6; − 8;
9
51
Without using a calculator, rank all the numbers in ascending order.

For more exercises, visit www.everythingmaths.co.za and click on ’Practise Maths’.

1a. 2DDW 1b. 2DDX 1c. 2DDY 1d. 2DDZ 1e. 2DF2 1f. 2DF3 1g. 2DF4 1h. 2DF5
1i. 2DF6 1j. 2DF7 2a. 2DF8 2b. 2DF9 2c. 2DFB 2d. 2DFC 3. 2DFD

1.1.6 Products EMAB

Mathematical expressions are just like sentences and their parts have special names. You should be familiar
with the following words used to describe the parts of mathematical expressions.

3x2 + 7xy − 53

Name Examples
term 3x2 ; 7xy ; −53
expression 3x2 + 7xy − 53
coefficient 3; 7
exponent 2; 1; 3
base x; y; 5
constant 3; 7; 5
variable x; y
equation 3x2 + 7xy − 53 = 0

Multiplying a monomial and a binomial EMAC

A monomial is an expression with one term, for example, 3x or y 2 . A binomial is an expression with two terms,
for example, ax + b or cx + d.

Worked example 7: Simplifying brackets

QUESTION

Simplify: ( )
2a (a − 1) − 3 a2 − 1

12 1.1.6. Products
13
SOLUTION

( ) ( )
2a (a − 1) − 3 a2 − 1 = 2a (a) + 2a (−1) + (−3) a2 + (−3) (−1)
= 2a2 − 2a − 3a2 + 3
= −a2 − 2a + 3

Multiplying two binomials EMAD

Here we multiply (or expand) two linear binomials:

(ax + b)(cx + d)

(ax + b) (cx + d) = (ax) (cx) + (ax) d + b (cx) + bd

= acx2 + adx + bcx + bd
= acx2 + x (ad + bc) + bd

Worked example 8: Multiplying two binomials

QUESTION

Find the product: (3x − 2) (5x + 8)

SOLUTION

(3x − 2) (5x + 8) = (3x) (5x) + (3x) (8) + (−2) (5x) + (−2) (8)
= 15x2 + 24x − 10x − 16
= 15x2 + 14x − 16

The product of two identical binomials is known as the square of the binomial and is written as:
2
(ax + b) = a2 x2 + 2abx + b2

If the two terms are of the form ax + b and ax − b then their product is:

(ax + b) (ax − b) = a2 x2 − b2

This product yields the difference of two squares.

Multiplying a binomial and a trinomial EMAF

A trinomial is an expression with three terms, for example, ax2 + bx + c. Now we can learn how to multiply a
binomial and a trinomial.

M : 1. Algebraic expressions 13
14
To find the product of a binomial and a trinomial, multiply out the brackets:
(A + B) (C + D + E) = A (C + D + E) + B (C + D + E)

Worked example 9: Multiplying a binomial and a trinomial

QUESTION
( )
Find the product: (x − 1) x2 − 2x + 1

SOLUTION

Step 1: Expand the bracket

( ) ( ) ( )
(x − 1) x2 − 2x + 1 = x x2 − 2x + 1 − 1 x2 − 2x + 1 = x3 − 2x2 + x − x2 + 2x − 1

Step 2: Simplify
( )
(x − 1) x2 − 2x + 1 = x3 − 3x2 + 3x − 1

Exercise 1 – 4:

1. Expand the following products:

a) 2y(y + 4) b) (y + 5)(y + 2) c) (2 − t)(1 − 2t)
d) (x − 4)(x + 4) e) −(4 − x)(x + 4) f) −(a + b)(b − a)
g) (2p + 9)(3p + 1) h) (3k − 2)(k + 6) i) (s + 6)2
j) −(7 − x)(7 + x) k) (3x − 1)(3x + 1) l) (7k + 2)(3 − 2k)
m) (1 − 4x)2 n) (−3 − y)(5 − y) o) (8 − x)(8 + x)
p) (9 + x) 2
q) (−7y + 11)(−12y + 3) r) (g − 5)2
s) (d + 9)2 t) (6d + 7)(6d − 7) u) (5z + 1)(5z − 1)
v) (1 − 3h)(1 + 3h) w) (2p + 3)(2p + 2) x) (8a + 4)(a + 7)
y) (5r + 4)(2r + 4) z) (w + 1)(w − 1)

2. Expand the following products:

a) (g + 11)(g − 11) b) (4b − 2)(2b − 4) c) (4b − 3)(2b − 1)

d) (6x − 4)(3x + 6) e) (3w − 2)(2w + 7) f) (2t − 3)2

14 1.1.6 Products
15
 g) (5p − 8)2 h) (4y + 5)2 i) (2y 6 + 3y 5 )(−5y − 12)
j) 9(8y 2 − 2y + 3) k) (−2y 2 − 4y + 11)(5y − 12) l) (7y 2 − 6y − 8)(−2y + 2)
m) (10y + 3)(−2y 2 − 11y + 2) n) (−12y − 3)(2y 2 − 11y + 3) o) (−10)(2y 2 + 8y + 3)
p) (7y + 3)(7y 2 + 3y + 10) q) (a + 2b)(a2 + b2 + 2ab) r) (x + y)(x2 − xy + y 2 )
s) 3m(9m2 + 2) + 5m2 (5m + 6) t) 4x2 (10x3 + 4) + 4x3 (2x2 + 6) u) 3k 3 (k 2 + 3) + 2k 2 (6k 3 + 7)
v) (3x + 2)(3x − 2)(9x2 − 4) w) (−6y 4 + 11y 2 + 3y)(y + 4)(y − 4)
x) (x + 2)(x − 3)(x2 + 2x − 3) y) (a + 2)2 − (2a − 4)2

3. Expand the following products:

a) (2x + 3)2 − (x − 2)2 b) (2a2 − a − 1)(a2 + 3a + 2)

c) (y 2 + 4y − 1)(1 − 4y − y 2 ) d) 2(x − 2y)(x2 + xy + y 2 )

e) 3(a − 3b)(a2 + 3ab − b2 ) f) (2a − b)(2a + b)(2a2 − 3ab + b2 )

g) 2(3x + y)(3x − y) − (3x − y)2 h) (x + y)(x − 3y) + (2x − y)2

( )( ) ( )( )
x 3 x 4 2 x 4
i) − + j) x − +
3 x 4 x x 3 x
1 1 1
k) (10x − 12y) + 13 (15x − 18y) l) a(4a + 6b) + (8a + 12b)
2 2 4

4. What is the value of b, in (x + b)(x − 1) = x2 + 3x − 4

5. What is the value of g, in (x − 2)(x + g) = x2 − 6x + 8
6. In (x − 4)(x + k) = x2 + bx + c:

a) For which of these values of k will b be positive?

−3; −1; 0; 3; 5
b) For which of these values of k will c be positive?
−3; −1; 0; 3; 5
c) For what real values of k will c be positive?
d) For what real values of k will b be positive?

7. Answer the following:

( )2
4
a) Expand x + .
x
( )2
4 16
b) Given that x + = 14, determine the value of x2 + 2 without solving for x.
x x
8. Answer the following:
( )2
1
a) Expand: a +
a
( ) ( )2
1 1
b) Given that a + = 3, determine the value of a + without solving for a.
a a
( ) ( )2
1 1
c) Given that a − = 3, determine the value of a + without solving for a.
a a

M:1 Algebraic expressions 15

16
 9. Answer the following:
( )2
1
a) Expand: 3y +
2y
( )2
1 1
b) Given that 3y + = 4, determine the value of 3y + without solving for y.
2y 2y
10. Answer the following:
( )2
1
a) Expand: a +
3a
( )( )
1 1 1
b) Expand: a + a2 − + 2
3a 3 9a
1 1
c) Given that a + = 2, determine the value of a3 + without solving for a.
3a 27a3

For more exercises, visit www.everythingmaths.co.za and click on ’Practise Maths’.

1a. 2DFG 1b. 2DFH 1c. 2DFJ 1d. 2DFK 1e. 2DFM 1f. 2DFN
1g. 2DFP 1h. 2DFQ 1i. 2DFR 1j. 2DFS 1k. 2DFT 1l. 2DFV
1m. 2DFW 1n. 2DFX 1o. 2DFY 1p. 2DFZ 1q. 2DG3 1r. 2DG4
1s. 2DG5 1t. 2DG6 1u. 2DG7 1v. 2DG8 1w. 2DG9 1x. 2DGB
1y. 2DGC 1z. 2DGD 2a. 2DGF 2b. 2DGH 2c. 2DGJ 2d. 2DGK
2e. 2DGM 2f. 2DGN 2g. 2DGP 2h. 2DGQ 2i. 2DG2 2j. 2DGR
2k. 2DGS 2l. 2DGT 2m. 2DGV 2n. 2DGW 2o. 2DGX 2p. 2DGY
2q. 2DGZ 2r. 2DH2 2s. 2DH3 2t. 2DH4 2u. 2DH5 2v. 2DH6
2w. 2DH7 2x. 2DH8 2y. 2DH9 3a. 2DHB 3b. 2DHC 3c. 2DHD
3d. 2DHF 3e. 2DHG 3f. 2DHH 3g. 2DHJ 3h. 2DHK 3i. 2DHM
3j. 2DHN 3k. 2DHP 3l. 2DHQ 4. 2DHR 5. 2DHS 6. 2DHT
7. 2DHV 8. 2DHW 9. 2DHX 10. 2DHY

1.1.7 Factorisation EMAG

Factorisation is the opposite process of expanding brackets. For example, expanding brackets would require
2(x + 1) to be written as 2x + 2. Factorisation would be to start with 2x + 2 and end up with 2(x + 1).

factorising

2(x + 1) 2x + 2

expanding

The two expressions 2(x + 1) and 2x + 2 are equivalent; they have the same value for all values of x.

In previous grades, we factorised by taking out a common factor and using difference of squares.

16 1.1.7 Factorisation
17
Common factors EMAH

Factorising based on common factors relies on there being factors common to all the terms.

For example, 2x − 6x2 can be factorised as follows:

2x − 6x2 = 2x(1 − 3x)

And 2(x − 1) − a(x − 1) can be factorised as follows:

(x − 1)(2 − a)

Worked example 10: Factorising using a switch around in brackets

QUESTION

Factorise:
5(a − 2) − b(2 − a)

SOLUTION

Use a “switch around” strategy to find the common factor.

Notice that 2 − a = −(a − 2)

5(a − 2) − b(2 − a) = 5(a − 2) − [−b(a − 2)]

= 5(a − 2) + b(a − 2)
= (a − 2)(5 + b)

Exercise 1 – 5:

Factorise:

1. 12x + 32y 2. −2ab2 − 4a2 b 3. 18ab − 3bc

4. 12kj + 18kq 5. −12a + 24a3 6. −2ab − 8a
7. 24kj − 16k 2 j 8. −a2 b − b2 a 9. 72b2 q − 18b3 q 2
10. 125x6 − 5y 2 11. 6x2 + 2x + 10x3 12. 2xy 2 + xy 2 z + 3xy
13. 12k 2 j + 24k 2 j 2 14. 3a2 + 6a − 18 15. 7a + 4

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1. 2DJ2 2. 2DJ3 3. 2DJ4 4. 2DJ5 5. 2DJ6 6. 2DJ7 7. 2DJ8 8. 2DJ9
9. 2DJB 10. 2DJC 11. 2DJD 12. 2DJF 13. 2DJG 14. 2DJH 15. 2DJJ

M:1 Algebraic expressions 17

18
Difference of two squares EMAJ

We have seen that (ax + b)(ax − b) can be expanded to a2 x2 − b2 .

Therefore a2 x2 − b2 can be factorised as (ax + b)(ax − b).

For example, x2 − 16 can be written as x2 − 42 which is a difference of two squares. Therefore, the factors of
x2 − 16 are (x − 4) and (x + 4).

To spot a difference of two squares, look for expressions:

• consisting of two terms;

• with terms that have different signs (one positive, one negative);
• with each term a perfect square.

For example: a2 − 1; 4x2 − y 2 ; −49 + p4 .

Worked example 11: The difference of two squares

QUESTION

Factorise: 3a(a2 − 4) − 7(a2 − 4).

SOLUTION

Step 1: Take out the common factor (a2 − 4)

3a(a2 − 4) − 7(a2 − 4) = (a2 − 4)(3a − 7)

Step 2: Factorise the difference of two squares (a2 − 4)

(a2 − 4)(3a − 7) = (a − 2)(a + 2)(3a − 7)

Exercise 1 – 6:

Factorise:

1. 4(y − 3) + k(3 − y) 2. a2 (a − 1) − 25(a − 1) 3. bm(b + 4) − 6m(b + 4)

4. a2 (a + 7) + 9(a + 7) 5. 3b(b − 4) − 7(4 − b) 6. 3g(z + 6) + 2(6 + z)
7. 4b(y + 2) + 5(2 + y) 8. 3d(r + 5) + 14(5 + r) 9. (6x + y)2 − 9

18 1.1.7 Factorisation
19
10. 4x2 − (4x − 3y)2 11. 16a2 − (3b + 4c)2 12. (b − 4)2 − 9(b − 5)2
13. 4(a − 3)2 − 49(4a − 5) 14. 16k 2 − 4 15. a2 b2 c2 − 1
1 1 2
16. a2 − 4b2 17. x − 2 18. y 2 − 8
9 2
19. y 2 − 13 20. a2 (a − 2ab − 15b2 ) − 9b2 (a2 − 2ab − 15b2 )

For more exercises, visit www.everythingmaths.co.za and click on ’Practise Maths’.

1. 2DJM 2. 2DJN 3. 2DJP 4. 2DJQ 5. 2DJR 6. 2DJS 7. 2DJT 8. 2DJV
9. 2DJW 10. 2DJX 11. 2DJY 12. 2DJZ 13. 2DK2 14. 2DK3 15. 2DK4 16. 2DK5
17. 2DK6 18. 2DK7 19. 2DK8 20. 2DK9

Factorising by grouping in pairs EMAK

The taking out of common factors is the starting point in all factorisation problems. We know that the factors
of 3x + 3 are 3 and (x + 1). Similarly, the factors of 2x2 + 2x are 2x and (x + 1). Therefore, if we have an
expression:
2x2 + 2x + 3x + 3

there is no common factor to all four terms, but we can factorise as follows:
( 2 )
2x + 2x + (3x + 3) = 2x (x + 1) + 3 (x + 1)

We can see that there is another common factor (x + 1). Therefore, we can write:

(x + 1) (2x + 3)

We get this by taking out the (x + 1) and seeing what is left over. We have 2x from the first group and +3
from the second group. This is called factorising by grouping.

Worked example 12: Factorising by grouping in pairs

QUESTION

Find the factors of 7x + 14y + bx + 2by.

SOLUTION

Step 1: There are no factors common to all terms

Step 2: Group terms with common factors together
7 is a common factor of the first two terms and b is a common factor of the second two terms. We see that the
ratio of the coefficients 7 : 14 is the same as b : 2b.

7x + 14y + bx + 2by = (7x + 14y) + (bx + 2by)

= 7 (x + 2y) + b (x + 2y)

M : 1. Algebraic expressions 19
20
Step 3: Take out the common factor (x + 2y)

7 (x + 2y) + b (x + 2y) = (x + 2y) (7 + b)

OR

Step 4: Group terms with common factors together

x is a common factor of the first and third terms and 2y is a common factor of the second and fourth terms
(7 : b = 14 : 2b).

Step 5: Rearrange the equation with grouped terms together

7x + 14y + bx + 2by = (7x + bx) + (14y + 2by)

= x (7 + b) + 2y (7 + b)

Step 6: Take out the common factor (7 + b)

x (7 + b) + 2y (7 + b) = (7 + b) (x + 2y)

Step 7: Write the final answer

The factors of 7x + 14y + bx + 2by are (7 + b) and (x + 2y).

Exercise 1 – 7:

Factorise the following:

1. 6d − 9r + 2t5 d − 3t5 r 2. 9z − 18m + b3 z − 2b3 m 3. 35z − 10y + 7c5 z − 2c5 y

4. 6x + a + 2ax + 3 5. x2 − 6x + 5x − 30 6. 5x + 10y − ax − 2ay
7. a2 − 2a − ax + 2x 8. 5xy − 3y + 10x − 6 9. ab − a2 − a + b
10. 14m − 4n + 7jm − 2jn 11. 28r − 20x + 7gr − 5gx 12. 25d − 15m + 5yd − 3ym
13. 45q − 18z + 5cq − 2cz 14. 6j − 15v + 2yj − 5yv 15. 16a − 40k + 2za − 5zk
16. ax − bx + ay − by + 2a − 2b 17. 3ax + bx − 3ay − by − 9a − 3b

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1. 2DKB 2. 2DKC 3. 2DKD 4. 2DKF 5. 2DKG 6. 2DKH 7. 2DKJ
8. 2DKK 9. 2DKM 10. 2DKN 11. 2DKP 12. 2DKQ 13. 2DKR 14. 2DKS
15. 2DKT 16. 2DKV 17. 2DKW

20 1.1.7 Factorisation
21
Factorising a quadratic trinomial EMAM

Factorising is the reverse of calculating the product of factors. In order to factorise a quadratic, we need to find
the factors which, when multiplied together, equal the original quadratic.

Consider a quadratic expression of the form ax2 + bx. We see here that x is a common factor in both terms.
Therefore ax2 + bx factorises as x (ax + b). For example, 8y 2 + 4y factorises as 4y (2y + 1).

Another type of quadratic is made up of the difference of squares. We know that:

(a + b) (a − b) = a2 − b2

So a2 − b2 can be written in factorised form as (a + b) (a − b).

This means that if we ever come across a quadratic that is made up of a difference of squares, we can immedi-
ately write down the factors. These types of quadratics are very simple to factorise. However, many quadratics
do not fall into these categories and we need a more general method to factorise quadratics.

We can learn about factorising quadratics by looking at the opposite process, where two binomials are multi-
plied to get a quadratic. For example:
(x + 2) (x + 3) = x2 + 3x + 2x + 6
= x2 + 5x + 6

We see that the x2 term in the quadratic is the product of the x-terms in each bracket. Similarly, the 6 in the
quadratic is the product of the 2 and 3 in the brackets. Finally, the middle term is the sum of two terms.

So, how do we use this information to factorise the quadratic?

Let us start with factorising x2 + 5x + 6 and see if we can decide upon some general rules. Firstly, write down
the two brackets with an x in each bracket and space for the remaining terms.
(x ) (x )

Next, decide upon the factors of 6. Since the 6 is positive, possible combinations are: 1 and 6, 2 and 3, −1
and −6 or −2 and −3

Therefore, we have four possibilities:

Option 1 Option 2 Option 3 Option 4

(x + 1) (x + 6) (x − 1) (x − 6) (x + 2) (x + 3) (x − 2) (x − 3)

Next, we expand each set of brackets to see which option gives us the correct middle term.

Option 1 Option 2 Option 3 Option 4

(x + 1) (x + 6) (x − 1) (x − 6) (x + 2) (x + 3) (x − 2) (x − 3)
x2 + 7x + 6 x2 − 7x + 6 x2 + 5x + 6 x2 − 5x + 6

We see that Option 3, (x + 2) (x + 3), is the correct solution.

The process of factorising a quadratic is mostly trial and error but there are some strategies that can be used to
ease the process.

M:1 Algebraic expressions 21

22
General procedure for factorising a trinomial EMAN

1. Take out any common factor in the coefficients of the terms of the expression to obtain an expression of
the form ax2 + bx + c where a, b and c have no common factors and a is positive.
2. Write down two brackets with an x in each bracket and space for the remaining terms: (x ) (x )
3. Write down a set of factors for a and c.
4. Write down a set of options for the possible factors for the quadratic using the factors of a and c.
5. Expand all options to see which one gives you the correct middle term bx.
IMPORTANT!

If c is positive, then the factors of c must be either both positive or both negative. If c is negative, it means only
one of the factors of c is negative, the other one being positive. Once you get an answer, always multiply out
your brackets again just to make sure it really works.

Worked example 13: Factorising a quadratic trinomial

QUESTION

Factorise: 3x2 + 2x − 1.

SOLUTION

Step 1: Check that the quadratic is in required form ax2 + bx + c

Step 2: Write down a set of factors for a and c

(x ) (x )
The possible factors for a are: 1 and 3

The possible factors for c are: −1 and 1

Write down a set of options for the possible factors of the quadratic using the factors of a and c. Therefore,
there are two possible options.
Option 1 Option 2
(x − 1) (3x + 1) (x + 1) (3x − 1)
3x2 − 2x − 1 3x2 + 2x − 1

Step 3: Check that the solution is correct by multiplying the factors

(x + 1) (3x − 1) = 3x2 − x + 3x − 1
= 3x2 + 2x − 1

Step 4: Write the final answer

3x2 + 2x − 1 = (x + 1) (3x − 1)

22 1.1.7 Factorisation
23
Exercise 1 – 8:

Factorise the following:

1. x2 + 8x + 15 2. x2 + 9x + 8 3. x2 + 12x + 36
4. 2h2 + 5h − 3 5. 3x2 + 4x + 1 6. 3s2 + s − 10
7. x2 − 2x − 15 8. x2 + 2x − 3 9. x2 + x − 20
10. x2 − x − 20 11. 2x2 − 22x + 20 12. 6a2 + 14a + 8
13. 6v 2 − 27v + 27 14. 6g 2 − 15g − 9 15. 3x2 + 19x + 6
16. 3x2 + 17x − 6 17. 7x2 − 6x − 1 18. 6x2 − 15x − 9
19. a2 − 7ab + 12b 20. 3a2 + 5ab − 12b2 21. 98x4 + 14x2 − 4
22. (x − 2)2 − 7(x − 2) + 12 23. (a − 2)2 − 4(a − 2) − 5 24. (y + 3)2 − 3(y + 3) − 18
25. 3(b2 + 5b) + 12 26. 6(a2 + 3a) − 168

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1. 2DKY 2. 2DKZ 3. 2DM2 4. 2DM3 5. 2DM4 6. 2DM5
7. 2DM6 8. 2DM7 9. 2DM8 10. 2DM9 11. 2DMB 12. 2DMC
13. 2DMD 14. 2DMF 15. 2DMG 16. 2DMH 17. 2DMJ 18. 2DMK
19. 2DMM 20. 2DMN 21. 2DMP 22. 2DMQ 23. 2DMR 24. 2DMS
25. 2DMT 26. 2DMV

Sum and difference of two cubes EMAP

We now look at two special results obtained from multiplying a binomial and a trinomial:

Sum of two cubes:

( ) ( ) ( )
(x + y) x2 − xy + y 2 = x x2 − xy + y 2 + y x2 − xy + y 2
[ ( ) ( )] [ ( ) ( )]
= x x2 + x (−xy) + x y 2 + y x2 + y (−xy) + y y 2
= x3 − x2 y + xy 2 + x2 y − xy 2 + y 3
= x3 + y 3

Difference of two cubes:

( ) ( ) ( )
(x − y) x2 + xy + y 2 = x x2 + xy + y 2 − y x2 + xy + y 2
[ ( ) ( )] [ ( ) ( )]
= x x2 + x (xy) + x y 2 − y x2 + y (xy) + y y 2
= x3 + x2 y + xy 2 − x2 y − xy 2 − y 3
= x3 − y 3

So we have seen that:

( )
x3 + y 3 = (x + y) x2 − xy + y 2
( )
x3 − y 3 = (x − y) x2 + xy + y 2

We use these two basic identities to factorise more complex examples.

M:1 Algebraic expressions 23

24
Worked example 14: Factorising a difference of two cubes

QUESTION

Factorise: a3 − 1.

SOLUTION

Step 1: Take the cube root of terms that are perfect cubes
( )
We are working with the difference of two cubes. We know that x3 − y 3 = (x − y) x2 + xy + y 2 , so we
need to identify x and y.
√
3
√
We start by noting that a3 = a and 3 1 = 1. These give the terms in the first bracket. This also tells us that
x = a and y = 1.

Step 2: Find the three terms in the second bracket

We can replace x and y in the factorised form of the expression for the difference of two cubes with a and 1.
Doing so we get the second bracket: ( ) ( )
a3 − 1 = (a − 1) a2 + a + 1

Step 3: Expand the brackets to check that the expression has been correctly factorised
( ) ( ) ( )
(a − 1) a2 + a + 1 = a a2 + a + 1 − 1 a2 + a + 1
= a3 + a2 + a − a2 − a − 1
= a3 − 1

Worked example 15: Factorising a sum of two cubes

QUESTION

Factorise: x3 + 8.

SOLUTION

Step 1: Take the cube root of terms that are perfect cubes
( )
We are working with the sum of two cubes. We know that x3 + y 3 = (x + y) x2 − xy + y 2 , so we need to
identify x and y.
√
3
√
We start by noting that x3 = x and 3 8 = 2. These give the terms in the first bracket. This also tells us that
x = x and y = 2.

Step 2: Find the three terms in the second bracket

We can replace x and y in the factorised form of the expression for the sum of two cubes with x and 2. Doing
so we get the second bracket: ( 3 ) ( )
x + 8 = (x + 2) x2 − 2x + 4

Step 3: Expand the brackets to check that the expression has been correctly factorised
( ) ( ) ( )
(x + 2) x2 − 2x + 4 = x x2 − 2x + 4 + 2 x2 − 2x + 4
= x3 − 2x2 + 4x + 2x2 − 4x + 8
= x3 + 8

24 1.1.7. Factorisation
25
Worked example 16: Factorising a difference of two cubes

QUESTION

Factorise: 16y 3 − 432.

SOLUTION

Step 1: Take out the common factor 16

( )
16y 3 − 432 = 16 y 3 − 27

Step 2: Take the cube root of terms that are perfect cubes
( )
We are working with the difference of two cubes. We know that x3 − y 3 = (x − y) x2 + xy + y 2 , so we
need to identify x and y.
√ √
We start by noting that 3
y 3 = y and 3 27 = 3. These give the terms in the first bracket. This also tells us that
x = y and y = 3.

Step 3: Find the three terms in the second bracket

We can replace x and y in the factorised form of the expression for the difference of two cubes with y and 3.
Doing so we get the second bracket:
( ) ( )
16 y 3 − 27 = 16 (y − 3) y 2 + 3y + 9

Step 4: Expand the brackets to check that the expression has been correctly factorised

16(y − 3)(y 2 + 3y + 9) = 16[(y(y 2 + 3y + 9) − 3(y 2 + 3y + 9)]

= 16[y 3 + 3y 2 + 9y − 3y 2 − 9y − 27]
= 16y 3 − 432

Worked example 17: Factorising a sum of two cubes

QUESTION

Factorise: 8t3 + 125p3 .

SOLUTION

Step 1: Take the cube root of terms that are perfect cubes
( )
We are working with the sum of two cubes. We know that x3 + y 3 = (x + y) x2 − xy + y 2 , so we need to
identify x and y.
√3
√
We start by noting that 8t3 = 2t and 3 125p3 = 5p. These give the terms in the first bracket. This also tells
us that x = 2t and y = 5p.

M : 1. Algebraic expressions 25
26
Step 2: Find the three terms in the second bracket
We can replace x and y in the factorised form of the expression for the difference of two cubes with 2t and 5p.
Doing so we get the second bracket:
( 3 ) [ ]
2 2
8t + 125p3 = (2t + 5p) (2t) − (2t) (5p) + (5p)
( )
= (2t + 5p) 4t2 − 10tp + 25p2

Step 3: Expand the brackets to check that the expression has been correctly factorised

( ) ( ) ( )
(2t + 5p) 4t2 − 10tp + 25p2 = 2t 4t2 − 10tp + 25p2 + 5p 4t2 − 10tp + 25p2
= 8t3 − 20pt2 + 50p2 t + 20pt2 − 50p2 t + 125p3
= 8t3 + 125p3

Exercise 1 – 9:

Factorise:
1. w3 − 8 2. g 3 + 64 3. h3 + 1

4. x3 + 8 5. 27 − m3 6. 2x3 − 2y 3

7. 3k 3 + 81q 3 8. 64t3 − 1 9. 64x2 − 1

10. 125x3 + 1 11. 25x3 + 1 12. z − 125z 4

13. 8m6 + n9 14. 216n3 − k 3 15. 125s3 + d3

16. 8k 3 + r3 17. 8j 3 k 3 l3 − b3 18. 27x3 y 3 + w3

1 27
19. 128m3 + 2f 3 20. p15 − y 12 21. − s3
8 t3
1 1
22. − h3 23. 72g 3 + v 3 24. 1 − (x − y)3
64q 3 3
25. h4 (8g 6 + h3 ) − (8g 6 + h3 ) 26. x(125w3 − h3 ) + y(125w3 − h3 )

27. x2 (27p3 + w3 ) − 5x(27p3 + w3 ) − 6(27p3 + w3 )

For more exercises, visit www.everythingmaths.co.za and click on ’Practise Maths’.

1. 2DMW 2. 2DMX 3. 2DMY 4. 2DMZ 5. 2DN2 6. 2DN3 7. 2DN4
8. 2DN5 9. 2DN6 10. 2DN7 11. 2DN8 12. 2DN9 13. 2DNB 14. 2DNC
15. 2DND 16. 2DNF 17. 2DNG 18. 2DNH 19. 2DNJ 20. 2DNK 21. 2DNM
22. 2DNN 23. 2DNP 24. 2DNQ 25. 2DNR 26. 2DNS 27. 2DNT

26 1.1.7. Factorisation
27
Step 3: Cancel the common factor in the numerator and the denominator to give the final answer

a+1
=
ax

Worked example 19: Simplifying fractions

QUESTION

Simplify:
x2 − x − 2 x2 + x
÷ , (x ̸= 0; x ̸= ±2)
x2 − 4 x2 + 2x

SOLUTION

Step 1: Factorise the numerator and denominator

(x + 1) (x − 2) x (x + 1)
= ÷
(x + 2) (x − 2) x (x + 2)

Step 2: Change the division sign and multiply by the reciprocal

(x + 1) (x − 2) x (x + 2)
= ×
(x + 2) (x − 2) x (x + 1)

Step 3: Write the final answer

=1

Worked example 20: Simplifying fractions

QUESTION

Simplify:
x−2 x2 x3 + x − 4
+ − , (x ̸= ±2)
x −4 x−2
2 x2 − 4

SOLUTION

Step 1: Factorise the denominators

x−2 x2 x3 + x − 4
+ −
(x + 2) (x − 2) x − 2 (x + 2) (x − 2)

28 1.1.8 Simplification of fractions

29
1.1.8 Simplification of fractions EMAQ

We have studied procedures for working with fractions in earlier grades.

a c ac
1. × = (b ̸= 0; d ̸= 0)
b d bd
a c a+c
2. + = (b ̸= 0)
b b b
a c a d ad
3. ÷ = × = (b ̸= 0; c ̸= 0; d ̸= 0)
b d b c bc

Note: dividing by a fraction is the same as multiplying by the reciprocal of the fraction.

In some cases of simplifying an algebraic expression, the expression will be a fraction. For example,

x2 + 3x
x+3
has a quadratic binomial in the numerator and a linear binomial in the denominator. We have to apply the
different factorisation methods in order to factorise the numerator and the denominator before we can simplify
the expression.

x2 + 3x x (x + 3)
=
x+3 x+3
=x (x ̸= −3)

If x = −3 then the denominator, x + 3 = 0 and the fraction is undefined.

Worked example 18: Simplifying fractions

QUESTION

Simplify:
ax − b + x − ab
, (x ̸= 0; x ̸= b)
ax2 − abx

SOLUTION

Step 1: Use grouping to factorise the numerator and take out the common factor ax in the denominator

(ax − ab) + (x − b) a (x − b) + (x − b)
=
ax − abx
2 ax (x − b)

Step 2: Take out common factor (x − b) in the numerator

(x − b) (a + 1)
=
ax (x − b)

M:1 Algebraic expressions 31

28
Step 2: Make all denominators the same so that we can add or subtract the fractions
The lowest common denominator is (x − 2) (x + 2).
( 2)
x−2 x (x + 2) x3 + x − 4
+ −
(x + 2) (x − 2) (x + 2) (x − 2) (x + 2) (x − 2)

Step 3: Write as one fraction

( ) ( )
x − 2 + x2 (x + 2) − x3 + x − 4
(x + 2) (x − 2)

Step 4: Simplify

x − 2 + x3 + 2x2 − x3 − x + 4 2x2 + 2
=
(x + 2) (x − 2) (x + 2) (x − 2)

Step 5: Take out the common factor and write the final answer
( )
2 x2 + 1
(x + 2) (x − 2)

Worked example 21: Simplifying fractions

QUESTION

Simplify:
2 x2 + x + 1 x
+ − 2 , (x ̸= 0; x ̸= ±1)
x2 − x x3 − 1 x −1

SOLUTION

Step 1: Factorise the numerator and denominator

( 2 )
2 x +x+1 x
+ −
x (x − 1) (x − 1) (x + x + 1) (x − 1) (x + 1)
2

Step 2: Simplify and find the common denominator

2 (x + 1) + x (x + 1) − x2
x (x − 1) (x + 1)

Step 3: Write the final answer

2x + 2 + x2 + x − x2 3x + 2
=
x (x − 1) (x + 1) x (x − 1) (x + 1)

M:1 Algebraic expressions 29

30
Exercise 1 – 10:

1. Simplify (assume all denominators are non-zero):

3a 2a + 10 5a + 20
a) b) c)
15 4 a+4
a2 − 4a 3a2 − 9a 9a + 27
d) e) f)
a−4 2a − 6 9a + 18
6ab + 2a 16x2 y − 8xy 4xyp − 8xp
g) h) i)
2b 12x − 6 12xy
9x2 − 16 b2 − 81a2 t 2 − s2
j) k) l)
6x − 8 18a − 2b s2 − 2st + t2
x2 − 2x − 15 x2 + 2x − 15 x2 − x − 6
m) n) 2 o)
5x − 25 x + 8x + 15 x3 − 27
a2 + 6a − 16 a2 − 4ab − 12b2 6a2 − 7a − 3
p) q) 2 r)
a3 − 8 a + 4ab + 4b2 3ab + b
2x2 − x − 1 qz + qr + 16z + 16r pz − pq + 5z − 5q
s) t) u)
x3 − x z+r z−q
hx − hg + 13x − 13g f 2 a − f a2
v) w)
x−g f −a
2. Simplify (assume all denominators are non-zero):

b2 + 10b + 21 2b2 + 14b x2 + 17x + 70 3x2 + 21x

a) ÷ b) ÷
3(b − 9)
2 30b2 − 90b 5(x − 100)
2 45x2 − 450x
z 2 + 17z + 66 2z 2 + 12z 3a + 9 7a + 21
c) ÷ d) ÷
3(z 2 − 121) 24z 2 − 264z 14 a+3
a2 − 5a 4a 3xp + 4p 12p2
e) × f) ÷
2a + 10 3a + 15 8p 3x + 4
24a − 8 9a − 3 a2 + 2a 2a + 4
g) ÷ h) ÷
12 6 5 20
p2 + pq 21q 5ab − 15b 6b2
i) × j) ÷
7p 8p + 8q 4a − 12 a+b
16 − x2 x+3 a3 + b3 5a + 5b
k) × l) × 2
x2 − x − 12 x + 4 a3 a + 2ab + b2
a−4 a2 + 2a + 1 3x + 2 x−2
m) × 2 n) × 2
a + 5a + 4 a − 3a − 4 x2 − 6x + 8 3x + 8x + 4
a2 − 2a + 8 a2 + a − 12 3 4x2 − 1 6x2 + 5x + 1 9x2 + 6x + 1
o) × − p) ÷ 2 ×
a2 + 6a + 8 3 2 3x2 + 10x + 3 4x + 7x − 3 8x2 − 6x + 1
x+4 x−2 p3 + q 3 3p − 3q
q) − r) × 2
3 2 p2 p − q2

3. Simplify (assume all denominators are non-zero):

x−3 x+5 2x − 4 x − 3
a) − b) − +1
3 4 9 4
3x − 4 x + 2 11 8
c) 1 + − d) +
4 3 a + 11 a − 8
12 6 12 8
e) − f) +
x − 12 x − 6 r + 12 r − 8

30 1.1.8 Simplification of fractions

31
 2 4 3 5 1
g) + + h) −
xy xz yz t−2 t−3
k+2 1 t+2 t+1
i) − j) +
k2 + 2 k + 2 3q 2q
3 2 x x2
k) + l) + 2
p2 − 4 (p − 2)2 x + y y − x2
1 3mn h 1
m) + n) − 2
m + n m 3 + n3 h3 − f 3 h + hf + f 2
x2 − 1 1 1 x2 − 2x + 1 x2 + x + 1
o) × − p) −
3 x−1 2 (x − 1)3 x3 − 1
1 2x t2 + 2t − 8 1 t+1
q) − 3 r) + 2 +
(x − 1)2 x −1 t2 + t − 6 t −9 t−3
x2 − 3x + 9 x−2 1 1 a2 + 2ab + b2 1
s) + 2 − t) + − 2
3
x + 27 x + 4x + 3 x − 2 a2 − 4ab + 4b 2 a − 8b
3 3 a − 4b2

4. What are the restrictions in the following:

1 3x − 9 3 1
a) b) c) − 2
x−2 4x + 4 x x −1

For more exercises, visit www.everythingmaths.co.za and click on ’Practise Maths’.

1a. 2DNW 1b. 2DNX 1c. 2DNY 1d. 2DNZ 1e. 2DP2 1f. 2DP3 1g. 2DP4 1h. 2DP5
1i. 2DP6 1j. 2DP7 1k. 2DP8 1l. 2DP9 1m. 2DPB 1n. 2DPC 1o. 2DPD 1p. 2DPF
1q. 2DPG 1r. 2DPH 1s. 2DPJ 1t. 2DPK 1u. 2DPM 1v. 2DPN 1w. 2DPP 2a. 2DPQ
2b. 2DPR 2c. 2DPS 2d. 2DPT 2e. 2DPV 2f. 2DPW 2g. 2DPX 2h. 2DPY 2i. 2DPZ
2j. 2DQ2 2k. 2DQ3 2l. 2DQ4 2m. 2DQ5 2n. 2DQ6 2o. 2DQ7 2p. 2DQ8 2q. 2DQ9
2r. 2DQB 3a. 2DQC 3b. 2DQD 3c. 2DQF 3d. 2DQG 3e. 2DQH 3f. 2DQJ 3g. 2DQK
3h. 2DQM 3i. 2DQN 3j. 2DQP 3k. 2DQQ 3l. 2DQR 3m. 2DQS 3n. 2DQT 3o. 2DQV
3p. 2DQW 3q. 2DQX 3r. 2DQY 3s. 2DQZ 3t. 2DR2 4a. 2DR3 4b. 2DR4 4c. 2DR5

1.1.9 Chapter summary EMAR

• – N: natural numbers are {1; 2; 3; . . .}

– N0 : whole numbers are {0; 1; 2; 3; . . .}
– Z: integers are {. . . ; −3; −2; −1; 0; 1; 2; 3; . . .}
• A rational number is any number that can be written as a
b where a and b are integers and b ̸= 0.
• The following are rational numbers:
– Fractions with both numerator and denominator as integers
– Integers
– Decimal numbers that terminate
– Decimal numbers that recur (repeat)

M:1 Algebraic expressions 31

32
• Irrational numbers are numbers that cannot be written as a fraction with the numerator and denominator
as integers.
• If the nth root of a number cannot be simplified to a rational number, it is called a surd.
√ √
• If a and b are positive whole numbers, and a < b, then n a < n b.
• A binomial is an expression with two terms.
• The product of two identical binomials is known as the square of the binomial.
• We get the difference of two squares when we multiply (ax + b) (ax − b)
• Factorising is the opposite process of expanding the brackets.
• The product of a binomial and a trinomial is:

(A + B) (C + D + E) = A (C + D + E) + B (C + D + E)
• Taking out a common factor is the basic factorisation method.
• We often need to use grouping to factorise polynomials.
• To factorise a quadratic we find the two binomials that were multiplied together to give the quadratic.
• The sum of two cubes can be factorised as:
( )
x3 + y 3 = (x + y) x2 − xy + y 2

• The difference of two cubes can be factorised as:

( )
x3 − y 3 = (x − y) x2 + xy + y 2

• We can simplify fractions by incorporating the methods we have learnt to factorise expressions.
• Only factors can be cancelled out in fractions, never terms.
• To add or subtract fractions, the denominators of all the fractions must be the same.

32 1.1.9 Chapter summary

33
Solutions to exercises
1 Algebraic expressions
Exercise 1 – 1:

1. a) Z b) rational b) 3
25
b) (ii) c) rational c) 29
50
2. a) between the rectangle and Z d) irrational d) 2589
10 000
b) (ii) 6. a) rational
10. a) 0,1̇
3. a) real b) rational
b) 0,12
b) real c) irrational
c) 0,123
c) non-real d) rational
d) 0,114145
d) undefined 7. a) 7 and 11
√ 11. a) 0,5̇
e) non-real b) −
√ 8 ; 3,3231089... ; 3 +
b) 0,5̇
f) real 2; π
√ c) 0,2̇1̇
4. a) rational c) −1
d) 0,6̇
b) irrational d) −3 ; 0 ; −8 45 ; 22
7 ; 7 ; 1,34
7
; 9 10 ; 11 e) 1,27
c) rational, an integer, a whole num-
e) −3 ; 7 ; 11 f) 4,83̇
ber and a natural number
f) 14 g) 2,1̇
d) irrational 0
5
8. a) (i) 555 (ii) rational 12. a)
e) irrational 9
57
f) irrational b) (i) any three digits (ii) irrational b) 90
c) (i) any three digits (ii) irrational 4
g) rational, an integer, a whole num- c) 9
ber and a natural number d) (i) any three digits (ii) irrational 526
d) 99
h) irrational e) (i) 545 (ii) rational 163
e) 33
i) irrational 9. a) 1
10 130
f) 33
5. a) rational

Exercise 1 – 2:

1. a) 12,566 2. a) 345,0440 3. a) 9,87 c) 648 768,22

b) 3,317 b) 1361,73 b) 4,93 5. 11
c) 0,267 c) 728,009052 c) 14,80 6. R 5,03
d) 1,913 d) 0,0370 d) 14,8044...
e) 6,325 e) 0,45455 4. a) 0,01
f) 0,056 f) 0,08 b) 519 854,59

Exercise 1 – 3:

1. a) 4 and 5 e) 12 and 13 i) 4 and 5 c) 3,9

b) 5 and 6 f) 7 and 8 j) 4 and 5 d) 3,5
√ √
c) 1 and 2 g) 8 and 9 2. a) 3,1 3. − 8 ; − 94 ; 0,45 ; 0,45 ;
d) 4 and 5 h) 4 and 5 b) 9,1 27
√ √
7 ; 19 ; 6 ; 2π ; 51

Exercise 1 – 4:

1. a) 2y 2 + 8y t) 36d2 − 49 m) −20y 3 − 116y 2 − 13y + 6

b) y 2 + 7y + 10 u) 25z 2 − 1 n) −24y 3 + 126y 2 − 3y − 9
c) 2t − 5t + 2
2
v) 1 − 9h 2
o) −20y 2 − 80y − 30
d) x2 − 16 w) 4p2 + 10p + 6 p) 49y 3 + 42y 2 + 79y + 30
e) x2 − 16 x) 8a2 + 60a + 28 q) a3 + 4a2 b + 5ab2 + 2b3
f) a − b
2 2 2
y) 10r + 28r + 16 r) x3 + y 3
2
g) 6p + 29p + 9 z) w − 1
2
s) 52m3 + 6m + 30m2
h) 3k2 + 16k − 12 2. a) g 2 − 121 t) 48x5 + 16x2 + 24x3
i) s2 + 12s + 36 b) 8b2 − 20b + 8 u) 15k5 + 9k3 + 14k2
j) x2 − 49 c) 8b2 − 10b + 3 v) 81x4 − 72x + 16
k) 9x − 1 2
d) 18x + 24x − 24
2
w) −6y 6 + 107y 4 + 3y 3 − 176y 2 −
l) −14k2 + 17k + 6 e) 6w2 + 17w − 14 48y

m) 16x2 − 8x + 1 f) 4t2 − 12t + 9 x) x4 + x3 − 11x2 − 9x + 18

n) y − 2y − 15
2
g) 25p − 80p + 64
2 y) −3a2 + 20a − 12
o) −x2 + 64 h) 16y 2 + 40y + 25 3. a) 3x2 + 16x + 5

p) x2 + 18x + 81 i) −10y 7 − 39y 6 − 36y 5 b) 2a4 + 5a3 − 5a − 2

q) 84y − 153y + 33
2
j) 72y − 18y + 27
2 c) −y 4 − 8y 3 − 14y 2 + 8y − 1

r) g − 10g + 25
2
k) −10y + 4y + 103y − 132
3 2 d) 2x3 − 2x2 y − 2xy 2 − 2y 3

s) d2 + 18d + 81 l) −14y 3 + 26y 2 + 4y − 16 e) 3a3 − 30ab2 + 9b3

Solutions
 f) 8a4 −12a3 b+2a2 b2 +3ab3 −b4 5. −4 b) 9
g) 9x2 + 6xy − 3y 2 6. a) 5 c) 13
h) 5x2 − 6xy − 2y 2 b) k = −3 or k = −1 9. a) 9y 2 + 3 + 1
4y 2
x2 7 3 c) k < 0
i) 12 + 12 + x2
b) 16
d) k > 4 10. a) a2 + 2
+ 1
x2
j) + 10
− 8 3 9a2
3 3 x2 7. a) x2 + 8 + 16
x2 3 1
k) 10x − 12y b) a + 27a3
b) 6
l) 2a2 + 3ab + 2a + 3b c) 6
8. a) a2 + 2 + 1
a2
4. 4

Exercise 1 – 5:

1. 4(3x + 8y) 6. −2a(b + 4) 11. 2x(3x + 1 + 5x2 )

2. −2ab(b + 2a) 7. 8kj(3 − 2k) 12. xy(2y + yz + 3)
3. 3b(6a − c) 8. −ab(a + b) 13. 12k2 j(1 + 2j)
4. 6k(2j + 3q) 9. 18b q(4 − bq)
2
14. 3(a2 + 2a − 6)
5. 12a(−1 + 2a ) 2
10. 5(5x − y)(5x + y)
3 3
15. 7a + 4

Exercise 1 – 6:

1. (y − 3)(4 − k) 8. (r + 5)(3d + 14) 15. (abc − 1)(abc + 1)

(1 )(1 )
2. (a − 1)(a − 5)(a + 5) 9. (6x + y − 3)(6x + y + 3) 16. 3 a − 2b 3 a + 2b
( )( )
3. (b + 4)(m)(b − 6) 10. 3(2x − y)(3y − 2x) 17. 2 12 x + 1 12 x − 1
4. (a + 7)(a2 + 9) 11. (4a + 3b + 4c)(4a − 3b − 4c) √ √
18. (y − 8)(y + 8)
5. (b − 4)(3b + 7) 12. (−2b + 11)(4b − 19) √ √
19. (y − 13)(y + 13)
6. (z + 6)(3g + 2) 13. (29 − 26a)(30a − 41)
20. (a − 3b)(a − 5b)(a + 3b)2
7. (y + 2)(4b + 5) 14. (4k − 2)(4k + 2)

Exercise 1 – 7:

1. (2d − 3r)(3 + t5 ) 7. (a − x)(a − 2) 13. (5q − 2z)(9 + c)

2. (z − 2m)(9 + b3 ) 8. (y + 2)(5x − 3) 14. (2j − 5v)(3 + y)
3. (7z − 2y)(5 + c5 ) 9. (−a + b)(a + 1) 15. (2a − 5k)(8 + z)
4. (3 + a)(2x + 1) 10. (7m − 2n)(2 + j) 16. (a − b)(x + y + 2)
5. (x + 5)(x − 6) 11. (7r − 5x)(4 + g) 17. (3a + b)(x − y − 3)
6. (5 − a)(x + 2y) 12. (5d − 3m)(5 + y)

Exercise 1 – 8:

1. (x + 5)(x + 3) 10. (x − 5)(x + 4) 19. (a − 4b)(a − 3b)

2. (x + 8)(x + 1) 11. 2(x + 1)(x + 10) 20. (3a − 4b)(a + 3b)
3. (x + 6)2 12. 2 (a + 1) (3a + 4) 21. 2((7x + 2)(7x − 1))
4. (h + 3)(2h − 1) 13. 3 (2v − 3) (v − 3) 22. (x − 6)(x − 5)
5. (x + 1)(3x + 1) 14. 3 (g − 3) (2g + 1) 23. (a − 7)(a − 1)
6. (s + 2)(3s − 5) 15. (3x + 1)(x + 6) 24. (y − 3)(y + 6)
7. (x + 3)(x − 5) 16. (3x − 1)(x + 6) 25. 3(b + 4)(b + 1)
8. (x + 3)(x − 1) 17. (7x + 1)(x − 1) 26. 6(a + 7)(a − 4)
9. (x + 5)(x − 4) 18. 3(2x + 1)(x − 3)

Exercise 1 – 9:

√
3 √
3 √
3
1. (w − 2)(w2 + 2w + 4) 11. ( 25x + 1)(( 25)2 x2 − 25x + 1) 21. ( 3t − s)( t92 + 3s
t + s2 )
2
2. (g + 4)(g − 4g + 16) 12. (z)(1 − 5z)(1 + 5z + 25z 2 ) 22. 1
( 4q − 1
h)( 16q + h
+ h2 )
2 4q
2
3. (h + 1)(h − h + 1) 13. (2m2 + n3 )(4m4 − 2m2 n3 + n6 )
23. 1
3 (6g + v)(36g 2 − 6gv + v 2 )
4. (x + 2)(x2 − 2x + 4) 14. (6n − k)(36n2 + 6nk + k2 )
15. (5s + d)(25s2 − 5sd + d2 ) 24. (1 − x + y)(1 − x + y + x2 − 2xy + y 2 )
5. (3 − m)(9 + 3m + m2 )
16. (2k + r)(4k − 2kr + r )
2 2 25. (h−1)(h+1)(h2 +1)(2g 2 +h)(4g 4 −
6. 2(x − y)(x2 + xy + y 2 ) 2g 2 h + h2 )
7. 3(k + 3q)(k2 − 3kq + 9q 2 ) 17. (2jkl − b)(4j k l + 2jklabc + b )
2 2 2 2
26. (x + y)(5w − h)(25w2 + 5wh + h2 )
8. (4t − 1)(16t2 + 4t + 1) 18. (3xy + w)(9x2 y 2 − 3xyw + w2 )
27. (x − 6)(x + 1)(3p + w)(9p2 − 3pw +
9. (8x − 1)(8x + 1) 19. 2(4m + f )(16m2 − 4mf + f 2 ) w2 )
( )( )
10. (5x + 1)(25x2 − 5x + 1) 20. p5 − 12 y 4 p10 + 12 p5 y 4 + 14 y 8

Solutions
 Exercise 1 – 10:

2(k+2)
1. a) a
5 q) a−6b
a+2b k) −1 i) (k2 +2)(k+2)
a+5 2a−3 5(a2 −ab+b2 )
b) 2 r) l) j) 5t+7
b a3 6q
c) 5 s) 2x+1
m) 1 5p−2
x(x+1) a+4 k) (p+2)(p−2)2
d) a
t) q + 16 n) 1
3a (x−4)(x+2) 2x2 −xy
e) 2 l)
u) p + 5 2a2 −14a+15
(x+y)(x−y)
f) a+3 o) 6 m+n
a+2 v) h + 13 m) m2 −mn+n2
a(3b+1) p) 1
g) b
w) af 14−x n) f
q) 6 (h+f )(h2 +hf +f 2 )
h) 4xy 2. a) 5
3 3(p2 −pq+q 2 ) 2x−1
r) o)
p(y−2) b) 3 p2
6
i) 3y p) 0
c) 4 3. a) x−27
j) 3x+4 12 −x2 +3x+1
2 d)
3(a+3)
47−x
q) (x−1)2 (x2 +x+1)
98 b)
k) − b+9
2 4a2 (a−5)
36
2t2 +5t−8
e) 6(a+5)2 c) 5x−8 r) t2 −9
t+s 12
l) t−s 19a x2 −9x−1
(3x+4)2 d)
x+3 f) (a+11)(a−8) s)
m) 5 96p2 (x+3)(x+1)(x−2)
6x
x−3 4(3a−1) e) (x−12)(x−6) a2 +4b−4b2
n) x+3
g) 3(a−1)
t) (a−2b)2 (a+2b)
20r
f)
o) x+2 h) 2a (r+12)(r−8)
4. a) x ̸= 2
x2 +3x+9 2z+4y+3x
a+8 i) 3q g) xyz b) x ̸= −1
p) 8
a2 +2a+4
j) 30b3 h) 4t−12
(t−2)(t−3)
c) x ̸= 0 and x ̸= ±1
4(a+b)

Exercise 1 – 11:

1. a) between the rectangle and Z d) 1,00 22. a) a2 + 10a + 25

b) (i) 12. a) 3,142 b) n2 + 24n + 144
2. a) non-real b) 1,618 c) d2 − 8d + 16
b) undefined c) 1,414 d) 49w2 − 4
c) real d) 2,718 e) 144q 2 − 1
d) real 13. 1523,0020 f) x2 + 4x + 4
e) non-real 14. 1982,940290 g) 25k2 − 16
f) real 15. 101,5238 h) 10f 2 + 18f + 8
3. a) Irrational 16. a) 1,414 i) 18n2 + 51n + 30
b) Irrational b) 1,732 j) 2g 2 + 18g + 36
c) Rational c) 2,236 k) 16y 2 + 36y + 8
d) Rational d) 2,449 l) 7d2 − 19d − 6
4. a) rational 17. a) Irrational number. m) 6z 2 − 16z + 8
b) rational b) Irrational number. n) 25w2 − 110w + 121
c) irrational c) Rational number. o) 25s2 − 10s + 1
d) irrational d) Rational number. p) 9d2 − 48d + 64
√
5. a) −24 e) Rational number. q) 36f 3 + 25f 2 + 49f
√ √
3 √
b) − 24 ; 32 ; π 2 ; 26 ; π ; 39 ; f) Rational number. r) 74d4 + 16d + 24d2
√
7,1̇1̇ ; 7,12 ; 78 ; 9 ; 3π ; π 2 g) Irrational number. s) 59x3 + 10x2 + 49x
√ √3 √
c) − 24 ; π 2 ; 26 ; π ; 39 ; h) Rational number. 23. a) y 6 − y 5 + y 4 − 2y 3 − 7y 2 − 2y
√ 2
78 ; 3π ; π i) Rational number. b) 4x
3
d) 2 ; 7,1̇1̇ ; 7,12 ; 9 j) Irrational number. c) x4 − 2x2 + 1
e) 9 18. a) 0,71 d) 64a3 − 27b3
√
2
f) 0 b) 3,742 e) 2x3 + 4x2 y − 8xy 2 − 6y 3
6. a) 3
25
20. a) 2 and 3 f) 9a4 + 9a3 − 34a2 b2 + 25ab3 −
3 b) 3 and 4 25b4
b) 500
c) 4 and 5 g) y 2 − 1
y2
410
c) 99
d) 5 and 6 a2
d) 59
1 100 h) 9 − 3
a2
e) 1 and 2
e) 221 i) 6x
18 f) 2 and 3
37 j) −4x − 8
f) 45 g) 2 and 3
81
24. 4
g) 11 h) 3 and 4
25. a) −1 ; 0 ; 1 ; 6
8. a) 0,05̇ i) 9 and 10
b) 0 ; 1 ; 6
b) 1,5 j) 8 and 9
c) k > 0
78
9. k) 3 and 4
99 d) k > −2
10. a) (i) any three numbers (ii) irrational l) 4 and 5
26. a) 9a2 + 3 + 1
4a2
b) (i) 111 (ii) rational 21. a) 3,7
b) 27a3 − 1
8a3
11. a) 0,50 b) 10,5
b) 1,00 c) 6,9 c) 374 12
c) 0,11 d) 4,5 27. a) 64

Solutions
 b) 25 y) (2h − 5g)(9 + m3 ) f) 5
c) 2400800 z) (7d − 2s)(9 + u2 ) g) − 2+b
3
d) 100 30. a) 2 (a + 1) (3a + 4) h) 1
x−2
28. a) 11 × 13 b) 3 (g − 3) (2g + 1) x−7
i) 3
b) 23 × 3 × 7 c) (5g − r)(25g 2 + 5gr + r 2 )
j) 5
c) 29 × 31 d) (2r + z)(4r 2 − 2rz + z 2 ) a−5
k) (a+2)2
d) 32 × 11 e) (7m − 2n)(2 + j)
1
e) 3 × 13 × 41 f) (5d − 3m)(5 + y) l) a2 +6a+36

29. a) (a − 3)(a + 3) 2
g) (g − 3)(g +3g + 9) m) −2(2a + b)
b) 9(b − 3)(b + 3) h) (z+5)(z 2 − 5z + 25) n) s + 31
( )( )
c) m − 13 m + 1
3 i) 3(b − a)(3a − b) o) n + 8
d) 5(1 − ab3 )(1 + ab3 ) j) (4x + 5y)(y − 4x) p) p2 − 2pq + q 2
12−x2
e) b(2a − 3)(2a + 3)(4a2 + 9) k) 4(4x3 − 3y 4 )(4x3 + 3y 4 ) q) 6x
( )( )
f) (a − 5)(a − 5) l) 16 a − 12b2 a + 12b2 r) −14
(a+7)(a−7)
g) (4b + 7)(4b + 7) m) (a − 3)(2 − 9x)(2 + 9x)
32x3 +x+2
s)
h) −4b2 (6b3 − 1)2 n) (b + 3)(b − 10) 2x3
4a−1
i) (4 + x2 )(2 + x)(2 − x) o) (2x + 5y)(x + y) t) (2a+1)(2a−1)(a−1)
j) 7(x − 2)(x + y) p) (x − 5y)(x + 3y) u) 5x+20
6
k) (y − 10)(y + 3) q) (4x2 + 3)(x2 + 2) 4x3 +11x2 +7x+3
v)
l) (1 − x)2 (1 + x) r) 2(3x2 − 4)(x2 − 5) x(x+2)(x+3)

2(b2 +9)
m) (1 + p)(−2 + 3p) s) (3a + b)(3a − b)(x + y + 3) w) (b−3)(b+3)
n) x(x−2)+(1+y)(1−y)(1+y 2 ) t) 2(2y + 3)(y − 4) x(x+1)
x) x2 +x+6
o) (x − 1)(x2 + x + 1)(4b − x) u)
(x−4)(x−3)(x+3)
2 17z
p) (v − 7)(3m + 19) y) (z+12)(z−5)
v) (3rs − 1)(9r s + 3rs + 1)
2 2

)( 1 ) 7w
q) (3f + 19)(z + 3) ( 1 z) (w−11)(w−4)
p w) 5h +r − 5h
r
+ r2
r) 3(p − 1
3 )(p
2
+ 3 + 1
9)
25h2 32. (3x − 4)(x + 2)
s) (2x2 − 5y )(4x + 10x2 y 3 +
3 4 x) (j + k)(4n − b)(16n2 + 4nb + 33. 5x
25y 6 ) b2 )
34. 8,85
t) (−p)(12 + 18p + 7p2 ) 31. a) −8a + 4
35. a2 + 2ab + 4b2
(a+3b)2 (a−3b) b) 125a − 64b
3 3
u) 3
36. 9x2 − 3x + 1
c) 16m4 − 81
v) (2a − 5)(3a − 1) 37. a) x ̸= 1
3 and x ̸= −1
d) a2 + 4ab + 4b2 − c2
w) (s − 3)(s + 5) b) a ̸= b and a ̸= −3
e) 2
x) (2v + 3h)(8 + j 5 )

2 Exponents

500 Solutions