Lesson Notes By Weeks and Term - Junior Secondary School 2

SUBJECT: MATHEMATICS

CLASS:  JSS 2

DATE:

TERM: 1st TERM

 
WEEK EIGHT

TOPIC: ALGEBRAIC EXPRESSION

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

1. -5 x 2y
2. -3a x -6b
3. -14a/7
4. -1/3 of 36x2                                

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

1.  -14a/7 = (-14) x a = (-14/7) x a

                               +7

    = -2 x a = -2a

4)    -1/3 of 36x2 = (+36) x x2 = - (36/3) x x2

                (-3)   

        = -12x2

Evaluation

Simplify the following

1.-16x/8

2. (-1/10) of 100z
3. (-2x) x (-9y)

Removing brackets

Example

Remove brackets from the following

a.8 (2c + 3d)    (b) 4y (3x-5)   (c) (7a-2b) 3a

Solution

8(2c+3d) = 8 x 2c + 8 x 3d

      = 16c + 24d

b.4y(3x-5) = 4y x 3x – 4y x 5

      = 12xy – 20y

c.(7a-2b)3a = 7a x 3a – 2b x 3a

          =21a2 – 6ab

Evaluation

Remove brackets from the following

1.-5x(11x – 2y)

2.-p(p – 5q)

3.(2c + 8d)(-2)

Expanding algebraic expressions

The expression (a+b)(b-5) means (a+2) x (b-5)

The terms in the first bracket, (a+2), multiply each term in the second bracket, b-5.

Example

Expand the following

1. (a+b) (c+d)   
2. (6-x) (3+y)
3. (2p-3q) (5p-4)

Solution

a.(a+b)(c+d) = c(a+b) + d(a+b)

          = ac+bc+ad+bd

b.(6-x)(3+y) = 3(6-x) + y (6-x)

            = 18 -3x +6y – xy

c.(2p-3q)(5p-4) = 5p(2p – 3q)-4(2p-3q)

          = 10p2 – 15pq – 8p + 12q   

Evaluation

Expand the following

1. (3+d)(2+d)    (b) (3x+4)(x-2)        (c) (2h-k)(3h+2k)         (d) (7m-5n)(5m+3n)

Factorization of algebraic expression

Example:

Factorize the following

1. 12y + 8z     (b)  4n2 – 2n    (c) 24pq – 16p2

Solution

1. 12y +8z

The HCF of 12y and 8z is 4

12y +8z = 4(12y/4 + 8z/4)

    = 4(3y + 2z)

1. 4n2 – 2n 

The HCF of 4n2 and 2n is 2n

4n2 – 2n = 2n(4n2/2n – 2n/2n)

    = 2n (2n-1)

1. 24pq – 16p2

The HCF of 24pq and 16p2 is 8p

24pq – 16p2 = 8p(24pq/8p - 16p2/8p)

            = 8p(3q – 2p)

Evaluation

Factorize the following:

1. 2abx + 7acx    (b) 3d2e + 5d2
2. 12ax + 8bx

READING ASSIGNMENT 

New General Mathematics, UBE Edition, Chapter 1, pages 20-21

Essential Mathematics by A J S Oluwasanmi, Chapter 1, pages 1-4

WEEKEND ASSIGNMENT

1. Simplify (-6x) x (-x) =_____ a) 6x ( b) 6x2 (c) -6x (d) -6x2
2. Remove brackets from -3(12a – 5) a) 15-36a b) 15a-36 c) 15a + 36 d) 36a – 15
3. Expand (a+3)(a+4)  (a) a2+7a+12 (b) a2+12a+7 (c) a2+12a-7 (d) a2+7a-12
4. Factorize abc + abd (a) ab(c+d) (b) ac(b+d) (c) ad(b+c) (d)abc(c+d)
5. Factorize 5a2 + 2ax (a) a(5a+2x) (b) 5(2a2+2x) (c) a(5x+2ax) (d)a2(5+2x)

THEORY

1. Expand the following:
2. (p+2q)(p+3q)
3. (5r+2s)(3r+4s)

2. Factorize the following
3. -18fg – 12g
4. -5xy + 10y