For maths students

1. CORE MATHEMATICS
ALGEBRAIC EXPRESSION

2. OBJECTIVES
By the end of the lesson the student will be able to:
1.
2.
3.
4.
5.
6.
7.
Express statements in mathematical symbols
Add and subtract algebraic expressions
Multiply two binomial expression
Factorize algebraic expression
Apply difference of two squares to solve problem
Perform operations on simple algebraic fractions
Determine the conditions under which an algebraic fraction is undefine

3. Algebraic Expression
• In algebra, letters are used to stand for numbers.
• This letter are called variables, eg: x, y, s,t etc.
• Algebraic expression is a collection of letters and
numbers/symbols combined with at least one of the
arithmetic operations +, -, ×, and ÷
• Eg: 2𝑥 + 𝑦 − 3, 3𝑎𝑏𝑐2, etc

4. Addition (+) Subtraction (-)
Sum difference
Increase decrease
More than less than
Greater than reduce
Exceed minus
Total deduct
Together fewer
Combine diminished
Key words

5. Multiplication (×) Division (÷)
Product quotient
Of divide
Bracket share
Multiply/multiple per
Times equal group
Double average
Key words

6. Forming Algebraic Expressions
• Statements in words are often written as algebraic expressions in mathematics
• Any letter may be used for the unknown number, but a different letter must be
used for each different unknown.
Statements Expression
Nine increased by a number 𝑥 9 + 𝑥
Fourteen decreased by a number 𝑝 14 − 𝑝
Seven less than a number 𝑡 𝑡 − 7
The product of 9 and a number 𝑛 9𝑛
Thirty two divided by a number 𝑦 32 ÷ 𝑦

7. Forming Algebraic Expressions
Statement Expression
5 more than a number 𝑥 + 5
Five more than twice a number 2𝑥 + 5
The product of a number and 6 added to
10
6𝑦 + 10
Seven divided by twice a number 7 ÷ 2𝑛 or 7
2𝑛
Three times a number decreased by 11 3𝑥 − 11
Exampls
If 4 times a number is subtracted from 3 and the result is multiplied by 7
Solution
Let the number be 𝑥
7(3 − 4𝑥)

8. Addition and Subtraction
We can only add or subtract like terms to a single term
Example
Simplify the following expression
i. 5𝑥 + 4 − 9𝑦 + 3𝑥 + 2𝑦 − 7
ii. 4𝑥 + 2𝑦 + 3𝑥 + 5𝑦
iii. 4𝑥2𝑦 + 5𝑥𝑦2 + 3𝑥2𝑦 − 2𝑥𝑦2
iv. 𝑥2 + 𝑥 + 2𝑥2

9. Solution
iii.4𝑥2𝑦 + 5𝑥𝑦2 + 3𝑥2𝑦 − 2𝑥𝑦2
4𝑥2𝑦 + 3𝑥2𝑦 + 5𝑥𝑦2 − 2𝑥𝑦2
7𝑥2𝑦 + 3𝑥𝑦2
i. 5𝑥 + 4 − 9𝑦 + 3𝑥 + 2𝑦 − 7
Grouping like terms
5𝑥 + 3𝑥 − 9𝑦 + 2𝑦 + 4 − 7
Simplify like terms
8𝑥 − 7𝑦 − 3 iv. 𝑥2 + 𝑥 + 2𝑥2
𝑥2 + 2𝑥2 + 𝑥
3𝑥2 + 𝑥
ii. 4𝑥 + 2𝑦 + 3𝑥 + 5𝑦
4𝑥 + 3𝑥 + 2𝑦 + 5𝑦
7𝑥 + 7𝑦

10. Multiplication and Division
Simplify the following expression
i. 4𝑝 × 8𝑝2
ii. 16𝑎3𝑏2 ÷ 2𝑎𝑏
iii. −2𝑎 × 4𝑐 × 5𝑏
15𝑦3
iv. 3𝑦2
Solution
iii. −2𝑎 × 4𝑐 × 5𝑏 = −40𝑎𝑏𝑐
i. 4𝑝 × 8𝑝2 = 32𝑝3
ii. 16𝑎3𝑏2 ÷ 2𝑎𝑏 = 8𝑎2𝑏
3𝑦2
3
iv. 15𝑦
= 5𝑦

11. Removing Brackets
𝑎 b + 𝑐 − 𝑑 = 𝑎𝑏 + 𝑎𝑐 − 𝑎𝑑
Examples
Simplify the following expressions
i. −3(2𝑥 − 4)
ii. 12𝑏 − 5 + 2𝑏 − 7
iii. 4 𝑥 + 1 + 2(2𝑥 + 1)
iv. 2𝑎[ 𝑎 + 3𝑏 + 4 2𝑎 − 𝑏 ]

12. Solution
i. −3 2𝑥 − 4 = −6𝑥 + 12 iv. 2𝑎[ 𝑎 + 3𝑏 + 4 2𝑎 − 𝑏 ]
= 2𝑎[𝑎 + 3𝑏 + 8𝑎 − 4𝑏]
= 2𝑎[𝑎 + 8𝑎 + 3𝑏 − 4𝑏]
= 2𝑎[9𝑎 − 𝑏]
= 18𝑎2 − 2𝑎𝑏
ii. 12𝑏 − 5 + 2𝑏 − 7
= 12𝑏 − 5 − 2𝑏 − 7
= 12𝑏 − 2𝑏 − 5 − 7
= 10𝑏 − 12
iii. 4 𝑥 + 1 + 2(2𝑥 + 1)
= 4𝑥 + 4 + 4𝑥 + 2
= 4𝑥 + 4𝑥 + 2 + 4
= 8𝑥 + 6

13. Binomial Expressions
• A binomial is an expression of two terms separated by a plus (+) or a minus
(-).
Examples; (𝑎 + 3), (𝑥2 − 𝑦2), (2𝑥 + 3𝑦) and (3𝑥 − 2).
• The multiplication of two binomials requires the use of the distributive
property.
𝑎 + 𝑏 𝑐 + 𝑑 = 𝑎 𝑐 + 𝑑 + 𝑏(𝑐 + 𝑑)
= 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑

14. i. 𝑎 + 2
Solution
𝑎 + 3 = a a + 3 + 2(a + 3)
= 𝑎2 + 3𝑎 + 2𝑎 + 6
= 𝑎2 + 5𝑎 + 6
Example
Expand the following
i. (𝑎 + 2)(𝑎 + 3)
ii. (2𝑥 + 𝑦)2
iii. (3𝑥 − 5)(3𝑥 + 5)
iv. (4𝑎 + 3𝑏)(4𝑎 − 3𝑏) ii. (2𝑥 + 𝑦)2 = (2𝑥 + 𝑦)(2𝑥 + 𝑦)
= 2𝑥 2𝑥 + 𝑦 + 𝑦(2𝑥 + 𝑦)
= 4𝑥2 + 2𝑥𝑦 + 2𝑥𝑦 + 𝑦2
= 4𝑥2 + 4𝑥𝑦 + 𝑦2

15. Solution
iii. 3𝑥 − 5 3𝑥 + 5 = 3𝑥 3𝑥 + 5 − 5(3𝑥 + 5)
= 9𝑥2 + 15𝑥 − 15𝑥 − 25
= 9𝑥2 − 25
iv. 4𝑎 + 3𝑏 4𝑎 − 3𝑏 = 4𝑎 4𝑎 − 3𝑏 + 3𝑏(4𝑎 − 3𝑏)
16𝑎2 − 12𝑎𝑏 + 12𝑎𝑏 − 9𝑏2
= 16𝑎2 − 9𝑏2

16. Factorization
Type 1
Common factors
The general method used to factorize terms with common factors is to find the
factor common to all the terms in the expression and bring it outside the brackets.
Example
Factorize the following expression
i. 3𝑎𝑥 + 6𝑎𝑦
ii. 54 − 81𝑥
iii. 𝑥3 − 𝑥2𝑦
iv. 100𝑥 − 25𝑥2

17. Solution
i. 3𝑎𝑥 + 6𝑎𝑦 = 3𝑎(𝑥 + 𝑦)
ii. 54 − 81𝑥 = 9(6 − 9𝑥)
iii. 𝑥3 − 𝑥2𝑦 = 𝑥2(𝑥 − 𝑦)
iv. 100𝑥 − 25𝑥2 = 25𝑥(4 − 𝑥)
Type 2
Method of Grouping
Examples
Factorize completely the following expression
i. 𝑚𝑥 − 𝑚𝑎 + 𝑛𝑥 − 𝑛𝑎
ii. 3𝑝2 + 2𝑞𝑡 + 6𝑞𝑝 + 𝑝𝑡
iii. 𝑦 6𝑥 − 2 − 𝑧(3𝑥 − 1)

18. Solution
i. 𝑚𝑥 − 𝑚𝑎 + 𝑛𝑥 − 𝑛𝑎 = 𝑚 𝑥 − 𝑎 + 𝑛(𝑥 − 𝑎)
= (𝑚 + 𝑛)(𝑥 − 𝑎)
ii. 3𝑝2 + 2𝑞𝑡 + 6𝑞𝑝 + 𝑝𝑡 = 3𝑝2 + 6𝑞𝑝 + 𝑝𝑡 + 2𝑞𝑡
= 3𝑝 𝑝 + 2𝑞 + 𝑡(𝑝 + 2𝑞)
= (3𝑝 + 𝑡)(𝑝 + 2𝑞)
iii. 𝑦 6𝑥 − 2 − 𝑧 3𝑥 − 1 = 2𝑦 3𝑥 − 1 − 𝑧(3𝑥 − 1)
= (2𝑦 − 𝑧)(3𝑥 − 1)
Note: the expressions in the brackets must be the same after bringing the common
factor out.

19. Factorization
Type 3
Quadratic Trinomials
• A trinomial is an expression containing three terms
• A quadratic trinomial has 3 terms i.e an 𝑥2 term, an 𝑥 term and a constant
term.
• An expression of the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, where a, b and c are constants not
equal to zero is a quadratic trinomial
Steps for factorizing quadratic trinomial
Step 1: find the product 𝒂𝒄 i.e multiply the coefficient of 𝑥2 by the constant.
Step 2: find the factors of the result in step 1 which can added or subtracted to
get the coefficient of 𝑥

20. Factorization
Step 3: replace the coefficient of 𝑥 (i.e 𝑏) by the factors obtained in step 2 and use
the method of grouping to complete the factorization.
Examples
Factorize the following
i. 2𝑥2 − 3𝑥 + 1
ii. 𝑥2 + 𝑥 − 6
iii. 3𝑝2 + 7𝑝𝑞 − 6𝑞2
iv. 𝑥2 − 15𝑥𝑦 + 56𝑦2

21. i. 2𝑥2 − 3𝑥 + 1
Solution
2𝑥2 − 3𝑥 + 1 = 2𝑥2 − 2𝑥 − 𝑥 + 1
= 2𝑥2 − 2𝑥
= 2𝑥 𝑥 − 1
− (𝑥 + 1)
− 1(𝑥 − 1)
= (2𝑥 − 1)(𝑥 − 1)
ii. 𝑥2 + 𝑥 − 6 = 𝑥2 + 3𝑥 − 2𝑥 − 6
= (𝑥2 + 3𝑥) − (2𝑥 − 6)
= 𝑥 𝑥 + 3 − 2(𝑥 + 3)
= (𝑥 − 2)(𝑥 + 3)
Step 1: the coefficient of 𝑥2 is 2 and
the constant is 1, there 2 × 1 = 2
Step 2: the possible pair of factors of
2 are (-1,-2) and (1,2) but only (-1,-2)
gives us -3 when added (i.e the
Coefficient of 𝑥)
Step 3: replace −3𝑥 by −2𝑥 − 𝑥 in
the given expression and factorize
using the grouping method

22. iii. 3𝑝2 + 7𝑝𝑞 − 6𝑞2
= 3𝑝2 + 9𝑝𝑞 − 2𝑝𝑞 − 6𝑞2
= (3𝑝2 + 9𝑝𝑞) − (2𝑝𝑞 − 6𝑞2)
= 3𝑝 𝑝 + 3𝑞 − 2𝑞(𝑝 + 3𝑞)
= (3𝑝 − 2𝑞)(𝑝 + 3𝑞)
Solution
iv. 𝑥2 − 15𝑥𝑦 + 56𝑦2
𝑥2 − 7𝑥𝑦 − 8𝑥𝑦 + 56𝑦2
(𝑥2 − 7𝑥𝑦) − (8𝑥𝑦 + 56𝑦2)
𝑥 𝑥 − 7𝑦 − 8𝑦(𝑥 − 7𝑦)
(𝑥 − 8𝑦)(𝑥 − 7𝑦)
Difference of Two Squares
Any expression of the form 𝑎2 − 𝑏2 is called difference of two squares.
Example
Factorize the following
i. 𝑥2 − 36𝑦2

23. ii. 4𝑎2 − 49
iii. 𝑥2 − 1
Solution
i. 𝑥2 − 36𝑦2 = 𝑥2 − 62𝑦2
= (𝑥 − 6𝑦)(𝑥 + 6𝑦)
ii. 4𝑎2 − 49 = 22𝑎2 − 72
= (2𝑎 − 7)(2𝑎 + 7)
iii. 𝑥2 − 1 = 𝑥2 − 12
= (𝑥 − 1)(𝑥 + 1)

24. Operations on algebraic fraction
Multiplication and Division
• To multiply algebraic fractions, multiply the numerators and denominators
separately. Example, 𝑎
× =
𝑥 𝑎𝑥
𝑏 𝑦 𝑏𝑦
• To divide algebraic fractions, multiply the expression by the inverse of the
algebraic fraction. Example, 𝑎
÷
𝑥
𝑏 𝑦
= × =
𝑎 𝑦 𝑎𝑦
𝑏 𝑥 𝑏𝑥
• Note: always check that your answer is n the simplest form.
Example
Simplify the following
i. 12𝑥𝑦
× 14𝑥
iii. 12𝑥𝑦
÷ 14𝑥
ii.
7 20
3𝑎𝑏
× 10𝑑4
15𝑐2𝑑3 9𝑎2
7 20
2 2
iv. 4𝑎 +8𝑎𝑏
÷ 5𝑎𝑏+10𝑏
3 9

25. Solution
2
i. 12𝑥𝑦
× 14𝑥
= 6𝑥 𝑦
iii. 12𝑥𝑦
÷ 14𝑥
= 12𝑥𝑦
× 20
= 120𝑦
ii.
7 20 5
3𝑎𝑏
× 10𝑑4
=
15𝑐2𝑑3 9𝑎2
2𝑏𝑑
9𝑎𝑐2 3 9 3
7 20 7 14𝑥 49
2 2
iv. 4𝑎 +8𝑎𝑏
÷ 5𝑎𝑏+10𝑏
= 4𝑎(𝑎+2𝑏)
×
9
5𝑏(𝑎+2𝑏)
= 12𝑎
5𝑏
Addition and Subtraction
To add and subtract algebraic fractions:
• Find the LCM of the denominator.
• Express all fractions in terms of the lowest common denominator.
• Simply the numerator to obtain the numerator of the answer.

26. Examples
i. −
3 2
𝑥+1 𝑥−1
iii.
1
1−𝑥
+
2
1+𝑥
ii. 2𝑥−1
− 𝑥+3
3 2
Solution
3
𝑥+1
i. −
𝑥−1
2
= 3 𝑥−1 −2(𝑥+1)
(𝑥+1)(𝑥−1)
ii. 2𝑥−1
− 𝑥+3
= 2 2𝑥−1 −3(𝑥+3)
= 3𝑥−3−2𝑥−2
(𝑥+1)(𝑥−1)
3 2 6
= 4𝑥−2−3𝑥−9
6
= 3𝑥−2𝑥−3−2
(𝑥+1)(𝑥−1)
= 4𝑥−3𝑥−2−9
6
=
𝑥−5
(𝑥2−1)
= 𝑥−11
6

27. Solution
1
1−𝑥
iii. +
1+𝑥
2
= 1 1+𝑥 +2(1−𝑥)
(1−𝑥)(1+𝑥)
= 1+𝑥+2−2𝑥
(1−𝑥)(1+𝑥)
= 1+2+𝑥−2𝑥
=
(1−𝑥)(1+𝑥)
3−𝑥
(1−𝑥2)
Zero or Undefined Algebraic Fractions
𝑔(𝑥)
Case 1: the algebraic faction 𝑓(𝑥)
= 0, when 𝑓 𝑥 = 0
Example,
𝑥−1
2𝑥+1
= 0, when 𝑥 − 1 = 0 or 𝑥 = 1

28. 𝑔(𝑥)
Case 2: any algebraic fraction of the form 𝑓(𝑥)
is undefined when 𝑔 𝑥 = 0
Example,
𝑥−1
2𝑥+1
is undefined, when 2𝑥 + 1 = 0 or 𝑥 = − 1
2
𝑔(𝑥)
Case 3: the domain of any algebraic fraction of the form 𝑓(𝑥)
is the set of real
values of 𝑥 except 𝑔 𝑥 = 0
Example 1
Find the value(s) of 𝑥 which makes the fractions undefined
i.
1
𝑥+3
ii.
𝑥+2
(𝑥−2)(𝑥+1)

29. Solution
i. The fraction is undefined if 𝑥 + 3 = 0 or 𝑥 = −3
ii. The fraction is undefined if 𝑥 − 2 𝑥 + 1 = 0 ⇒ 𝑥 = 2 or −1
𝑥
2𝑥−1
.
𝑥 =
Example 2
Find the domain of the fraction
Solution
2𝑥 − 1 = 0
1
2
2
{𝑥: 𝑥𝜖𝑅, except 𝑥 = 1
}

30. Try
Example 1
Simplify the following
Example 4
a. find the domain of the fraction
2𝑥
𝑥−4
b. find the values of 𝑥 which makes the
fraction
1
𝑥2+3𝑥+2
not defined
Example 3
factorize the following
completely
a. 𝑎𝑏 − 𝑏2 + 5𝑎 − 5𝑏
b. 18𝑥2 − 33𝑥𝑦 + 5𝑦2
c. 49𝑎2 − 4𝑏2
a. 22𝑥 − 4𝑥𝑦 − 7𝑥 + 3𝑥𝑦
b. 𝑥3 + 𝑥2 + 4𝑥3 − 3𝑥2
c. 20𝑎 + 2𝑎𝑏 − 6𝑎 − 5𝑎𝑏
d. 64𝑥𝑦 ÷ 8𝑦
e. 2𝑎𝑏𝑐 × 8𝑎𝑏2
c. If a number is added to 2 times another
number and their is divided by 6. write an
algebraic expression for the statement.
Example 4
simplify
𝑥−2
+
2 𝑥
𝑥2−4
a.
b.
13𝑝2𝑞
6𝑥2𝑦
× 39𝑞2𝑟
÷ 𝑥2𝑝2
18𝑦2𝑧 𝑥𝑞𝑟
Example 2
Simplify the following
a. 4 𝑎 + 𝑏 − 2(2𝑎 + 3𝑏)
b. 7(3𝑥 − 4𝑦 + 2)
c. (3𝑥 + 𝑦)(𝑥 − 𝑦)