This is a courseware on Algebraic Expression intended for high school teachers and students. It covers the concept and basic operations on algebraic expressions. Contacts Details: Mobile: +233 248870038 Email: ddeynu@aims.edu.gh, kwabla1991@gmail.com

1. Algebraic Expression
By: David Kwabla Deynu
Nungua Senior High School
January 12, 2021

2. Definition
An algebraic expression in mathematics is an expression
which is made up of variables and constants, along with
at least one of the operations (addition, subtraction etc.).
Equal to sign (=) is NOT involved.
Numbers, operations and variables form Algebraic
Expressions.
Numbers and operations form Numerical
Expressions.

3. More about Algebraic Expression
Terms: The parts of the expression.
Eg. 2x + 3y − 7 has three terms.
Coefficients: The numerical factor of a term that
contains a variable.
Eg. From 2x + 3y − 7; 2 and 3 are coefficients
Variables: A symbol (usually a letter) that
represents one or more numbers.
Eg. From 2x + 3y − 7; x and y are variables.
Constants: A term with no variable.
Eg. From 2x + 3y − 7; 7 is a constant.

4. Illustrations
2x + y + 3 Algebraic Expression
3 + 9 Numerical Expression
2x + y + 3 = 0 Equation
4mn + 3m + 5mn − 7m Algebraic Expression

5. Types of Algebraic Expression
Monomial: Expression with only one term.
Eg. 3x
Binomial: Expression with only two terms.
Eg. 2x + 1
Polynomial: Expression with more than one term.
Eg. 2x + 5y + 1
Try: Write down two examples of each of the three types
mentioned above

6. Operations on algebraic expressions
Algebraic expressions can be simplified by using the
operations such as addition, subtraction, multiplication
and division. Let’s see how this is possible.

7. Addition and Subtraction
Here, we only add and or subtract like terms. Rewrite
the given expression so that like terms are next to each
other. This process is termed collecting/grouping like
terms. Let’s look at few examples

8. Worked Example 1
Simplify 14y − x + 2y − 6x − 5y
Solution
14y + 2y − 5y − x − 6x
= (14 + 2 − 5)y − (1 + 6)x Grouping like terms
= 11y − 7x Simplifying
∴ 14y − x + 2y − 6x − 5y = 11y − 7x

9. Worked Example 2
Simplify: 6k + km + 2k + 12km + 3k
Solution
6k + 2k + 3k + km + 12km
= (6 + 2 + 3)k + (1 + 12)km
= 11k + 13km

10. Worked Example 3
Simplify: 11v2
w − 2v2
+ 3v2
− 8v2
w − 2v2
Solution
11v2
w − 8v2
w + 3v2
− 2v2
− 2v2
= (11 − 8)v2
w + (3 − 2 − 2)v2
= 3v2
w − v2

11. Practice Examples
Simplify the following expressions:
1 4x2
y + 5xy2
+ 3x2
y − 2xy2
2 18d − 8e − d + 2e
3 14hk3
− 2kh3
− 2hk3
+ 3kh3
4 8yz + z + 2yz − 2z − 4z
5 2p − 3q + 3p + 5q

12. Multiplication and Division
Here, the numbers and the variables are grouped
together. Then the appropriate laws of indices are used
to simplify the expression. Note the laws:
ax
× ay
= ax+y
ax
÷ ay
= ax−y
an
× bn
= (a × b)n
= (ab)n

13. Worked Example 1
Simplify the expression: 2x2
× 3y2
× x2
Solution
2x2
× 3y2
× x2
=2 × 3 × x2
× x2
× y2
=6 × x2+2
× y2
=6 × x4
× y2
=6x4
y2

14. Worked Example 2
Simplify the expression: 2ax × 3by × 4ab
Solution
2ax × 3by × 4ab =2 × 3 × 4 × a × a × b × b × x × y
=24 × a1+1
× b1+1
× xy
=24 × a2
× b2
× xy
=24a2
b2
xy
=24(ab)2
xy

15. Worked Example 3
Simplify the expression: 18nm3
÷ 3nm
Solution
18nm3
÷ 3nm =
18nm3
3nm
=
18
3
×
n
n
×
m3
m
=6 × 1 × m3−1
=6 × 1 × m2
=6m2

16. Worked Example 4
Simplify the expression:
30x5
y3
−5x3y2
Solution
30x5
y3
−5x3y2
= −
30
5
×
x5
x3
×
y3
y2
= − 6 × x5−3
× y3−2
= − 6 × x2
× y
= − 6x2
y

17. Practice Exercise
Simplify the following Expressions:
3xy3
× 4x2
y (1)
2s × 3ab × sxy (2)
16m3
n2
÷ 2mn (3)
−12x3
y2
3xy
(4)
25abc ÷ (−5ab) (5)
2st3
× 4s4
t2
× 3st. (6)

18. Multiplication of Two Binomial
Given the binomials; (a + b) and (c + d), we wish to find
the product; (a + b)(c + d). To achieve this, we must
use the distributive property; thus multiply each term in
one bracket by each term in the other bracket as shown
below.
(a + b)(c + d) =a(c + d) + b(c + d)
=ac + ad + bc + bd
NB: The bracket (c + d) was removed by multiplying its
content by a and b.

19. More on Multiplication of Two Binomials
Note the following:
(a + b)(a − b) = a2
− b2
The difference of two squares
(a + b)2
= (a + b)(a + b) = a2
+ 2ab + b2
A perfect square
(a − b)2
= (a − b)(a − b) = a2
− 2ab + b2
A perfect square

20. Worked Example 1
Multiply (3x + y)(4x + 2y)
Solution
(3x + y)(4x + 2y) =3x(4x + 2y) + y(4x + 2y)
=12x2
+ 6xy + 4xy + 2y2
=12x2
+ 10xy + 2y2
∴ (3x + y)(4x + 2y) =12x2
+ 10xy + 2y2

21. Worked Example 2
Expand (4a − 3b)(a + 6b)
Solution
(4a − 3b)(a + 6b) =4a(a + 6b) − 3b(a + 6b)
=4a2
+ 24ab − 3ab − 18b2
=4a2
+ 21ab − 18b2
∴ (4a − 3b)(a + 6b) =4a2
+ 21ab − 18b2

22. Worked Example 3
Expand (x + 3)2
Solution
(x + 3)2
=(x + 3)(x + 3)
=x2
+ 3x + 3x + 9
=x2
+ 6x + 9
∴ (x + 3)2
= x2
+ 6x + 9

23. Worked Example 4
Expand and simplify (2x + y)(x − y) + (2x − y)(x + y)
Solution
(2x + y)(x − y) + (2x − y)(x + y)
2x(x − y) + y(x − y) + [2x(x + y) − y(x + y)]
2x2
− 2xy + xy − y2
+ [2x2
+ 2xy − xy − y2
]
2x2
+ 2x2
− 2xy + 2xy + xy − xy − y2
− y2
4x2
− y2
∴ (2x + y)(x − y) + (2x − y)(x + y) = 4x2
− y2

24. Worked Example 5
Expand and simplify (x + h)2
− (x − h)2
Solution
(x + h)2
− (x − h)2
(x + h)(x + h) − [(x − h)(x − h)]
x(x + h) + h(x + h) − [x(x − h) − h(x − h)]
x2
+ hx + hx + h2
− [x2
− hx − hx + h2
]
x2
+ 2hx + h2
− [x2
− 2hx + h2
]
x2
− x2
+ 2hx + 2hx + h2
− h2
4hx
∴ (x + h)2
− (x − h)2
= 4hx

25. Practice Exercise
Expand and simplify the following.
1 (a − 3)(a + 4)
2 (x + y)2
− (x − y)2
3 (10x + 5y)(u − 3v) + (x + y)(u − v)
4 (m − n)2
5 3m + (2m − p)
6 4(x + 1) − 2(4x − 4)
7 (9 + k)2