Back to: MATHEMATICS JSS 2
Welcome to Class !!
We are eager to have you join us !!
In today’s Mathematics class, We will be discussing Algebraic Expressions. We hope you enjoy the class!
Content
 Definition with examples
 Expansion of algebraic expression
 Factorization of simple algebraic expressions
Definition with examples
In algebra, letters stand for numbers. The numbers can be whole or fractional, positive or negative.
Example
Simplify the following
 5 x 2y
 3a x 6b
 14a/7
 1/3 of 36x^{2}
Solution
1) 5 x 2y = 5 x (+2) x y
= (5 x 2) x y = 10y
2) 3a x 6b = (3) x a (6) x b
= (3) x (6) x a x b = 18ab
3)
$\frac{\u201314a}{7}=\frac{(\u201314)\times a}{7}=\left(\u2013\frac{14}{7}\right)\times a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\u20132\times a=\u20132a$
4)
$\frac{\u20131}{3}of36{x}^{2}=\frac{\u20131}{3}\times 36{x}^{2}=\u2013\left(\frac{36}{3}\right)\times {x}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\u2013\left(12\right)\times {x}^{2}=\u201312{x}^{2}$1/3 of 36x^{2} = (+36) x x^{2} = – (36/3) x x^{2}
^{ } (3)
= 12x^{2}
Self Evaluation
Simplify the following
 16x/8
 (1/10) of 100z
 (2x) x (9y)
Removing brackets
Example
Remove brackets from the following
a. 8 (2c + 3d) (b) 4y (3x5) (c) (7a2b) 3a
Solution
(a) 8(2c+3d) = 8 x 2c + 8 x 3d
= 16c + 24d
(b) 4y(3x5) = 4y x 3x – 4y x 5
= 12xy – 20y
(c) (7a2b)3a = 7a x 3a – 2b x 3a
=21a^{2} – 6ab
Self Evaluation
Remove brackets from the following
 5x(11x – 2y)
 p(p – 5q)
 (2c + 8d)(2)
Expanding algebraic expressions
The expression (a+2) (b5) means (a+2) x (b5)
The terms in the first bracket, (a+2), multiply each term in the second bracket, b5.
Example
Expand the following

 (a+b) (c+d)
 (6x) (3+y)
 (2p3q) (5p4)
Solution
(1) (a+b)(c+d) = c(a+b) + d(a+b)
= ac + bc + ad + bd
(2) (6x)(3+y) = 3(6x) + y (6x)
= 18 3x +6y – xy
(3) (2p3q)(5p4) = 5p(2p – 3q)4(2p3q)
= 10p^{2} – 15pq – 8p + 12q
Self Evaluation
Expand the following
(a) (3+d)(2+d)
(b) (3x+4)(x2)
(c) (2hk)(3h+2k)
(d) (7m5n)(5m+3n)
Factorization of algebraic expression
Example:
Factorize the following
(a) 12y + 8z
(b) 4n^{2 }– 2n
(c) 24pq – 16p^{2}
Solution
(a) 12y +8z
The HCF of 12y and 8z is 4
12y +8z =
$4\left(\frac{12y}{4}+\frac{8z}{4}\right)$= 4(3y + 2z)
(b) 4n^{2} – 2n
The HCF of 4n^{2} and 2n is 2n
4n^{2} – 2n =
$2n\left(\frac{4{n}^{2}}{2n}\u2013\frac{2n}{2n}\right)$= 2n (2n1)
(c) 24pq – 16p^{2}
The HCF of 24pq and 16p^{2} is 8p
24pq – 16p^{2} =
$8p\left(\frac{24pq}{8p}\u2013\frac{16{p}^{2}}{8p}\right)$= 8p(3q – 2p)
Self Evaluation
Factorize the following:
 2abx + 7acx (b) 3d^{2}e + 5d^{2}
 12ax + 8bx
READING ASSIGNMENT
New General Mathematics, UBE Edition, Chapter 1, pages 2021
Essential Mathematics by A J S Oluwasanmi, Chapter 1, pages 14
WEEKEND ASSIGNMENT
 Simplify (6x) x (x) =_____ (A) 6x (b) 6x^{2} (C) 6x (D) 6x^{2}
 Remove brackets from 3(12a – 5) (A) 1536a (B) 15a36 (C) 15a + 36 (C) 36a – 15
 Expand (a+3)(a+4) (A) a^{2}+7a+12 (B) a^{2}+12a+7 (C) a^{2}+12a7 (D) a^{2}+7a12
 Factorize abc + and (A) ab(c+d) (B) ac (b+d) (C) ad(b+c) (D) abc(c+d)
 Factorize 5a^{2} + 2ax (A) a(5a+2x) (B) 5(2a^{2}+2x) (C) a(5x+2ax) (D) a^{2}(5+2x)
THEORY
 Expand the following:
 (p+2q) (p+3q)
 (5r+2s) (3r+4s)
 Factorize the following
 18fg – 12g
 5xy + 10y
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
In our next class, we will be talking about Algebraic Fractions. We are very much eager to meet you there.
LEARN TO CODE IN 8 WEEKS. Pay Only ₦25000 To Join Class💃
Access Fun Video Lessons to Pass WAEC, NECO, JAMB, POSTUTME in One Sitting💃