Introduction

Expansion of algebraic expressions

If the content of a bracket are to be multiplies by

any quantity and the bracket is removed each term

must be multiplied by that quantity.

Case 1 worked out examples

Remove the brackets in the following algebraic expressions

and simplify:

i). 5(2x +3)

ii). b(3x-2y)

iii). 4(2x-5y-3)

i). 5(2x +3)

= 5 x 2x + 5 x 3

= 10x + 15

ii). b(3x - 2y)

= bx3x - bx2y

= 3bx - 2by

iii). 4(2x - 5y - 3)

= 4 x 2x - 4 x 5y - 4 x 3

= 8x - 20y – 12

Case 2    :worked out examples

If there is a positive (+) sign just before a bracket the sign

of each term inside is unchanged when the bracket is removed. 

If there is a negative (-) sign just before a bracket the sign

of each term inside must be changed when the bracket is removed.

i). 2x + (y +2) = 2x + y +2

ii). 3a + (2a -5) = 3a + 2a - 5

= 5a -5

iii). 4b - 3 (b+2) = 4b - 3b - 6

iv). 4m - (4-m) = 4m - 4 + m

= 4m + m - 4

= 5m -4

i). 5(x +2) -2(3x -8)

= 5x - 10 - 6x + 16

= 5x - 6x + 10 + 16

= -1x + 26

= 26-1x

ii). 4(3a + 5b) + 3(2a +b)

= 12a + 20b + 6a + 3b

= 12a + 6a + 20b + 3b

= 18a + 23b

iii). ab(a-ab) - a(ab-a²)

a²b -a²b²-a²b+a³

a²b-a²b-a²b²+a³

= 0 -a²b²+a³

= -a²b²+a³

iv). 3(m+n) - 6(n-m)

= 3m + 3n - 6n + 6m

= 3m + 6m + 3n - 6n

= 9m -3n

v). 4(3a-3) + 3(a-1)

= 12a - 12 + 3a -3

= 15a-15

Factorization of algebraic expressions

Earlier on in this lesson you learnt that the process of

removing the brackets in an algebraic expression is

called expression. 

The opposite process of expansion is called factorization. 

It is to find factors of an expression that can be multiplied

to form the algebraic expression.

Case 3 =  :worked out examples

Factorize the following:

i). 3a +6

ii). a - 4ab

iii). 12p²-8pg

iv). 2a³+4a²b²-6ab

Solutions

i). 3a + 6

3(a+2)

ii). a-4ab

= a(1-4b)

iii). 12p² - 8pq

= 4p(3p - 2q)

iv). 2a³+ 4a²b² - 6ab

= 2a(a² +2ab² - 3b)

If the expression to be factorized has even numbers of terms:

1. factor out any common factor(s)

2. Group the terms in pairs such that there is a common factor(s) in

each pair

3. New factorize each of the pair of terms

Case iv

a). x² - xy + 6x - 6y

There is no common factor to all the terms. x² is paired with -xy,

since "x" is common and 6x with -6y since "6" is common

(x² - xy) + (6x - 6y)

Factorize the common factor in each pair

x(x-y) + 6(x-y)

(x-y)(x+6)

= (x-y) and (x + 6) are the factors

x²-xy+6x-6y=(x-y)(x+6)

2t - 2u + tz - zu

There is no common factors in all the terms

(2t - 2u) + (tz - zu)

2(t - u) + z(t - u)

(t - u)(2 + z)

cd - de + d² - ce

Rearranging the terms to have a common factor

cdfd²-e(c+d)

d(c+d) - e(c +d)

(c+d)(d-e)