The document discusses algebra and its basic concepts. It begins by introducing algebra and defining it as the aspect of mathematics involving the use of numbers and letters. It then provides examples of solving algebraic puzzles using letters to represent unknown numbers. The document goes on to define key algebraic terms like constants, variables, coefficients and terms. It also outlines fundamental algebraic rules like commutativity, associativity and distributivity. It discusses how to collect like terms and factorize expressions. Finally, it covers topics like expanding and simplifying brackets, fractions and factorizing expressions.

1. ALGEBRA
BY ADJEKUKOR, CYNTHIA UFUOMA
1

2. CONTENTS
Algebra 1
Algebra 2
Algebra 3
Algebra 4
Algebra 5
Algebra 6
2

3. ALGEBRA 1
3

4. Contents
Introduction
Solution to Puzzle
Definition of terms
Rules in Algebra
Collecting Like Terms
Similar Terms
4

5. INTRODUCTION
 Algebra is that aspect of mathematics that involves the use
of numbers and letters or letters only.
 Puzzle
 Think of a number
 Add 15 to it
 Double the result
 Add this to the number you first thought of
 Divide the result by 3
 Take away the number you first thought of
 The answer is 10 Why?
5

6. SOLUTION TO PUZZLE
Let letter ‘a’ be the number
(a + 15) × 2 = 2a + 30
2a + 30 + a = 3a + 30
(3a + 30) ⁄ 3 = 3a ⁄ 3 + 30 / 3
3a ⁄ 3 + 30 / 3 = a + 10
a + 10 − a = 10
6

7. DEFINITION OF TERMS
Constant: the letter ‘a’ used in the puzzle is a constant.
Variable: 2 + 3 = 3 + 2 Thus a + b = b + a.
Now ‘a’ and ‘b’ are variables because they represent any
two numbers.
Coefficient and Term: 5 × y = 5y. Now number ‘5’ which
is attached to letter ‘y’ is seen as the coefficient of y.
While y is the term.
7

8. RULES IN ALGEBRA
There are similar to the rules in arithmetic.
Commutativity: two numbers ‘a’ and ‘b’ can
be added or multiplied in any other without
affecting the results.
a + b = b + a & a × b = b × a
But a − b ≠ b − a unless a = b
 & a ⁄ b ≠ b ⁄ a unless b = a and neither equals zero.
8

9. Rules in Algebra
Associativity: The way in which the numbers a , b and c
are associated under addition or multiplication does
not affect the result.
 a + (b + c) = (a + b) + c = a + b + c
a × (b × c) = (a × b) × c = a × b × c
But for division and multiplication
x − (y − z) ≠ (x−y) − z unless z = 0
And x ÷ (y ÷ z) ≠ (x ÷ y) ÷ z unless z = 1 and y ≠ 0
9

10. Rules in Algebra
Distributivity: this states that multiplication are
distributed over addition and subtraction from both
left to right.
• For example:
x (y + z) = xy + xz (distributed from the left)
(x + y)z = xz + yz (distributed from the right)
x(y − z) = xy − xz (distributed from the left)
(x − y)z = xz − xy (distributed from the right)
• But division is distributed only over addition and
subtraction from the right and not from the left.
10

11. COLLECTING LIKE TERMS
 Terms which have the same variable are called like terms.
 And they can be collected together by addition or
subtraction.
 For example
 4x + 3y − 2z + 5y − 3x + 4z
 First, collect like terms and rearrange the operation.
 4x − 3x + 3y + 5y − 2z + 4z
 With these, you can simplify
 x + 8y + 2z
11

12. Collecting Like Terms
Simplify 4uv − 7uz − 6wz + 2uv + 3wz
Collect like terms.
Note the terms here are double letter and
include: uv, uz and wz.
4uv + 2uv − 7uz − 6wz + 3wz
6uv − 7uz −3wz
12

13. SIMILAR TERMS
Factorize
ab + bc
First question, what term is similar between ab
and bc? It is b. So bring out b to stand alone.
b (a + c)
In expanding the bracket, you will have b × a = ab
+ b × c = bc.
13

14. Similar Terms
Factorize 3pq − 3qr.
Bring out what is similar to stand alone.
3q (p − r)
Factorise 9st − 3sv − 6sw
3s (3t − v − 2w)
14

15. ALGEBRA 2
15

16. Contents
Expanding the Brackets
Nested Brackets
Multiplication of Algebraic Expressions
of a Single Variable
Division of One Expression by another
16

17. EXPANDING BRACKETS
In expanding brackets, you multiply each of the terms
enclosed in brackets by the term outside the bracket.
But if the term outside the bracket has a negative sign,
the signs of the terms enclosed in the bracket will
change.
3x (y − 2z)
(3x × y) − (3x × 2z)
3xy − 6xz
17

18. Expanding Brackets
What if you have a negative sign?
− 2y (2x − 4z)
First multiply through by 2y
− [ (2y × 2x) − (2y × 4z)]
− [4xy − 8yz]
Multiply through by the negative sign
−4xy + 8yz
18

19. NESTED BRACKETS
 When ever an algebraic expression contains brackets that
are nested within each other, you will start simplifying from
the innermost bracket.
 4 {2a + 3 [5 −2 (a−b)]}
 Start simplifying from innermost bracket
 4 {2a + 3[5 − 2a − 2b]}
 Next bracket
 4{2a + 15 − 6a − 6b}
 Next bracket
 8a + 60 − 24a − 24b
19

20. Nested Brackets
Collect like terms
8a − 24a + 60 − 24b
Simplify
−16a + 60 −24b
Rearrange
−16a −24b + 60
20

21. Nested Brackets
Simplify
8x (y − z) + 2y (7x + z)
Expand the bracket
(8xy − 8xz) + (14xy + 2yz)
Multiply the second bracket through by +
8xy − 8xz + 14xy + 2yz
21

22. Nested Brackets
Collect like terms
8xy + 14xy − 8xz + 2yz
Simplify
22xy − 8xz + 2yz
Factorise
2 (11xy − 4xz + yz)
It can still be factorised further
2[ x (11y − 4z) + yz}
22

23. Nested Brackets
(3a − b) (b − 3a) + b2
Expand the brackets
3a (b − 3a) −b (b − 3a) + b2
3ab − 9a2 − b2 + 3ab + b2
Collect like terms
3ab + 3ab − 9a2 − b2 + b2
6ab − 9a2
Factorise
3a (2b − 3a)
23

24. MULTIPLICATION OF ALGEBRAIC
EXPRESSIONS OF A SINGLE VARIABLE
 (x + 2) (x + 3)
Multiply (x + 3) by x and (x + 3) by +2
x (x + 3) +2 (x + 3)
x2 + 3x + 2x + 6
Simplify
x2 + 5x + 6
24

25. Multiplication of Algebraic Expressions
of a Single Variable
(2x + 5) (x2 + 3x + 4)
Multiply (x2 + 3x + 4) by 2x and (x2 + 3x + 4) by +5
2x (x2 + 3x + 4) +5 (x2 + 3x + 4)
Expand the bracket
2x3 + 6x2 + 8x + 5x2 + 15x + 20
Collect like terms
2x3 + 6x2 + 5x2 + 8x + 15x + 20
Simplify
2x3 + 11x2 + 23x + 20
25

26. DIVISION OF ONE EXPRESSION BY
ANOTHER
 (12x3 − 2x2 − 3x + 28) ÷ 3x + 4
 So, we are going to do long division
 3x+4 √ (12x3 − 2x2 − 3x + 28)
 We will only divide the first term of the numerator (12x3) by 3x. Thus
4x2. Then multiply 3x+4 by 4x2. it will give you 12x3 + 16x.
 3x+4 √ (12x3 − 2x2 − 3x + 28)
 −(12x3 + 16x2)
 The result will be −18x2 − 3x + 28
 Then we will continue to divide the first term of every results by 3x.
And then multiply 3x+4 by the result. Then do the normal operation
until there is nothing to divide.
 Therefore (12x3 − 2x2 − 3x + 28) ÷ 3x + 4 is 4x2 − 6x +7
26

27. ALGEBRA 3
ALGEBRAIC FRACTIONS
27

28. Contents
Addition and Subtraction of
Algebraic Fractions
Multiplication and Division of
Algebraic Fractions
28

29. ADDITION AND SUBTRACTION OF
ALGEBRAIC FRACTIONS
Simplify
 a⁄b − c⁄d2 + d⁄a
First, look for the LCM
The LCM is abd2
Divide the LCM (the new denominator) by the other
denominators (b, d2, & a).
Then multiply each of your results by the
corresponding numerator of the denominators.
(a2d2 − abc + bd3) / abd2
29

30. Addition and Subtraction of Algebraic
Fractions
 Simplify 2/(x+1) + 4/(x+2)
 First, look for the LCM which is (x+1) (x+2)
 Divide the LCM by the denominator (x+1). The answer is (x+2). Then (x+2)
multiplied by 2 is 2(x+2).
 For the second one divide the LCM by the denominator. The answer is (x+1).
Then (x+1) multiplied by 4 is 4(x+1).
 Add your two results together and divide by the denominator
 [2(x+2) + 4(x+1)] / (x+1) (x+2)
 Expand the brackets of the numerator and denominator
 (2x + 4 + 4x +4) / ( x2 +3x +2)
 Collect like terms at the numerator
 (2x + 4x + 4 + 4) / ( x2 +3x +2)
 (6x + 8 )/ (x2 +3x +2)
30

31. MULTIPLICATION AND DIVISION
 (a/b) × (c/d) = (ac) / (bd)
 (a/b) ÷ (c/d) = (a/b) × (d/c) = (ad) / (bc)
 Simplify (2a / 3b) ÷ (a2b / 6)
 Remember the switch in division
 (2a/3b) × (6/a2b)
 (2a × 6) / (3b × a2b)
 (12a) / (3a2b2)
 Note 3a can divide both the numerator and denominator. 12a / 3a =
4 and 3a2b2 divided by 3a is ab2
 Therefore, your answer will be 4 / ab2
31

32. ALGEBRA 4
32

33. Contents
Factorisation of Algebraic Expression
Useful Products of Two Simple
Factors
Examples
33

34. FACTORISATION OF ALGEBRAIC
EXPRESSION
Common factors
• Factorize 8x4y3 + 6x3y2
This can be written as
8xxxxyyy + 6xxxyy
 what is common?
2xxxyy (4xy +3)
This can be written as
2x3y2 (4xy +3)
34

35. Factorisation of Algebraic Expression
Common factors by grouping
• Factorize 2ac + 6bc + ad +3bd
Group the expression in two brackets
(2ac + 6bc) + (ad +3bd)
What is common in each bracket?
2c(a+3b) +d (a+3b)
They re two (a+3b), so you pick one and select the outer
terms as a group.
(2c+d) (a+3b)
35

36. USEFUL PRODUCTS OF TWO SIMPLE
FACTORS
(a+b) (a−b)
(a+b)2 = (a+b) (a+b) = a(a+b) + b(a+b)
• a2 + ab + ab + b2 = a2 +2ab + b2
(a−b)2 = (a−b) (a−b) = a(a−b) − b(a−b)
• a2 −ab −ab + b2 = a2 −2ab + b2
(a−b) (a+b) = a(a+b) − b(a+b)
• a2 + ab −ab − b2 = a2 − b2
36

37. Useful Products of Two Simple Factors
Remember that
(a+b)2 = a2 + 2ab +b2
(a−b)2 = a2 − 2ab +b2
(a+b) (a−b) = a2 − b2 Difference of two squares
37

38. EXAMPLES
 Factorize 4a2 − 12a + 9
 Present this as
 22a2 − 12a + 32
 which is
 (2a)2 − 12a + 32
 This can be represented to look like this
 a2 − 2ab +b2
 Note a = 2a and b = 3 and 2ab= 2(2a)(3)
 Therefore (2a)2 − 12a + 32 can be written as
 (2a)2 − 2(2a)(3) + 32
 Recall again (a−b)2 = a2 − 2ab +b2 and a = 2a while b=3.
 Thus (2a)2 − 2(2a)(3) + 32 = (2a−3)2
 That is 4a2 − 12a + 9 = (2a−3)2
38

39. Examples
Factorize 16x2 + 40xy + 25y2
This can be written as
42x2 + 40xy + 52y2
Which is (4x)2 + 40xy + (5y)2
This looks like a2 + 2ab +b2 where a= 4x b =5y then 2ab =
2(4x)(5y)
Thus (4x)2 + 40xy + (5y)2 can be written as
(4x)2 + 2(4x)(5y) + (5y)2
Recall again that (a+b)2 = a2 + 2ab + b2
Thus (4x)2 + 2(4x)(5y) + (5y)2 = (4x + 5y)2
39

40. Examples
Factorise 9a2 − 16x2
This can be written as
32a2 − 42x2 which can also be written as
(3a)2 − (4x)2 that is similar to a2 − b2 where a= 3a and
b= 4x
Recall (a+b) (a−b) = a2 − b2
Therefore (3a)2 − (4x)2 = (3a + 4x) (3a − 4x)
Thus 9a2 − 16x2 = (3a + 4x) (3a − 4x)
40

41. ALGEBRA 5
TRINOMIAL EXPRESSION
41

42. Contents
Trinomial Expression
Quadratic Expression as the Product of
Two Simple Factors
Factorisation of Quadratic Expression ax2
+ bx +c When a = 1
42

43. TRINOMIAL EXPRESSION
These types of expression are always in three
terms
They are mostly quadratic equation.
They take the form ax2 + bx + c
43

44. QUADRATIC EXPRESSION AS THE
PRODUCT OF TWO SIMPLE FACTORS
(x+g) (x+k) = x (x+k) + g (x+k)
x2 + kx + gx + gk
x2 + (k+g)x + gk
The coefficient of the middle term (k+g)x sum
of the constant g and k and the last term is the
product of the constants g and k.
44

45. Quadratic Expression as the Product of
Two Simple Factors
(x−g) (x−k) = x (x−k) − g (x−k)
x2 −kx − gx + gk
x2 −(k+g)x +gk
The coefficient of the middle term is minus the
sum of the two constants g and k while the last
term is the product of the two constants g and k
45

46. Quadratic Expression as the Product of
Two Simple Factors
(x+g) (x−k) = x (x−k) + g (x−k)
x2 −kx + gx − gk
 x2 + (g−k)x − gk
The coefficient of the middle term is the
difference of the two constants g and k while
the last term is minus the product of the two
constants g and k.
46

47. FACTORISATION OF QUADRATIC
EXPRESSION ax2 + bx +c WHEN a = 1
(x + f1) and (x + f2) are the factors of the quadratic
expression.
If c is positive
f1 and f2 are factors of c and bx + c have a sign of b. then
the sum of f1 and f2 is b.
If c is negative
f1 and f2 are factors of c and have opposite signs. Then the
numerical larger one have the sign of b and the difference
between f1 and f2 is b.
47

48. Factorisation of Quadratic Expression
ax2 + bx +c When a = 1
 Factorize x2 + 5x + 6
 Note C is positive. So therefore f1 and f2 are factors of c. what
are the factors of 6?
 1 and 6
 2 and 3
 The factors have the sign of b which is +. And the sum of f1 and
f2 is b.
 So 1+6 = 7 ≠ b but 2+3 = 5 = b
 the factors of the quadratic expression will be (x + f1) and (x + f2)
 Which is (x+2) and (x+3)
 That is x2 + 5x + 6 = (x+2) and (x+3)
48

49. Factorisation of Quadratic Expression
ax2 + bx +c When a = 1
 Factorize x2 − 9x + 20
 Note C is positive. So therefore f1 and f2 are factors of c. what are the
factors of 20?
 1 and 20
 2 and 10
 4 and 5
 The factors have the sign of b which is −. And the sum of f1 and f2 is b.
 −1 −20 = −21≠b , −2 −10 =−12≠b and −4−5 = −9=b
 the factors of the quadratic expression will be (x + f1) and (x + f2)
 Which is (x−4) and (x−5)
 That is x2 −9x +20 = (x−4) and (x−5)
49

50. Factorisation of Quadratic Expression
ax2 + bx +c When a = 1
Factorize x2 − 2x −35
f1 and f2 are factors of c and have opposite signs. Then the
numerical larger one have the sign of b and the difference
between f1 and f2 is b.
The factors of 35 are
1 and 35, 5 and 7
The numerical larger one which is 7 have the sign of b that
is the negative sign thus 5 has the positive sign.
Then −7 + 5 = −2 = b
Thus x2 − 2x −35 = (x + 5) (x−7)
50

51. Algebra 6
51

52. Contents
Factorisation of Quadratic Expression ax2
+ bx +c When a ≠ 1
Test for Simple Factors
52

53. FACTORISATION OF QUADRATIC
EXPRESSION ax2 + bx +c WHEN a ≠ 1
Obtain /ac/
If C is positive, we will select two factors of /ac/ whose
sum is equal to /b/. Both of these factors will carry the sign
of b.
If C is negative, we will select two factors of /ac/ which
differ by the values of /b/. The numerically larger one will
have the sign of b. the two factors will thus possess
opposite signs.
 then ax2 + bx +c will now be written as
 ax2 + f1x + f2x +c
 you will now factorise by grouping.
53

54. Factorisation of Quadratic Expression
ax2 + bx +c When a ≠ 1
 Factorise 6x2 + 11x + 3
 ax2 + bx +c
 a = 6, b = 11, c = 3
 Now /ac/ = /6 × 3/ = /18/ =18
 Since c is positive, we are interested in .two factors of /ac/ whose sum is equal
to /b/. And the two factors are 2 and 9.
 So therefore ax2 + bx +c will now be written as
 ax2 + f1x + f2x +c
 6x2 + 2x + 9x + 3
 Factorise by grouping
 (6x2 + 2x) + (9x + 3)
 2x (3x + 1) +3 (3x +1)
 (2x +3) (3x+1)
54

55. Factorisation of Quadratic Expression
ax2 + bx +c When a ≠ 1
 Factorise 8x2 + 18x −5
 ax2 + bx +c
 a = 8, b = 18, c = −5
 Now /ac/ = /8 × −5 / = /−40/ = 40
 Since C is negative, we will select two factors of /ac/ which differ by the values of /b/.
The numerically larger one will have the sign of b. And the two factors are + 20 and −2
 So therefore ax2 + bx +c will now be written as
 ax2 + f1x + f2x +c
 8x2 −2x +20x −5
 Factorise by grouping
 (8x2 −2x) +(20x −5)
 2x(4x − 1) +5 (4x − 1)
 (2x+5) (4x − 1)
55

56. TEST FOR SIMPLE FACTORS
Some quadratic equations are not capable of
being written as the product of simple factors.
i.e. factors where all the coefficients are integers.
To determine whether ax2 + bx +c can be
factorised into two simple factors.
We will first evaluate the expression (b2 −4ac).
56

57. Test for Simple Factors
If b2 −4ac is a perfect square; can be written as k2.
then ax2 + bx +c can be factorised into simple
factors.
But if b2 −4ac is not a perfect square; cannot be
written as k2.
then ax2 + bx +c cannot be factorised into simple
factors
57

58. Test for Simple Factors
Test for the simple factors of 3x2 − 4x +5 .
a=3 b = −4 c=5
Using b2 −4ac , we will have (−4)2 − 4 (3) (5)
(−4)2 − 4 × 3 × 5
16 − 60
−44
Note −44 is not a perfect square, thus the expression
3x2 − 4x +5 have no simple factors.
58

59. Test for Simple Factors
 Test for the simple factors of 2x2 +5x −3 .
 a=2 b = 5 c=−3
 Using b2 −4ac , we will have (5)2 − 4 (2) (−3)
 (5)2 − 4 × 2 × −3
 25 − 4(−6)
 25 + 24
 49
 Note 49 is a perfect square; can be written as 72.
 thus the expression 2x2 + 5x −3 can be factorised into simple
factors
59

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60

61. THANK
YOU
FOR
LISTENING
61