Form 2 Unit 4 Lesson 1 – Factorization of Algebraic Expressions

Objectives

At the end of the lesson, students should be able to:

Find the highest common factor of two or more algebraic terms.

Factorise algebraic expressions.

Factors of any number are defined as the numbers which can divide that number evenly without leaving any remainder. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12.

Any number can be expressed as a product of its factors as explained below:

12 = 2 × 6

12 = 3 × 4

12 = 1 × 12

In terms of its prime factors 12 can be expressed as:

12 = 2 × 3 × 2

In algebra, when we factorise, the highest possible factor is brought out from each of the algebraic terms and used to divide each term. The remainder of the expression is put in a bracket.

Consider the algebraic expression –3y(y – 6).

If we expand this we will obtain –3y² + 18y. If we simply reverse the process and find the HCF of the terms of this expression i.e. HCF of –3y² and 18y we get 3y. Now this 3y is common to –3y² + 18y. Write 3y in front of the bracket and divide each term–3y² + 18y by 3y.Write the remainder inside the brackets.

⇒–3y² + 18y = 3y × –y + 3y × 6y

⇒–3y² + 18y = 3y(–y + 6y)

NOTE: FACTORIZATION IS THE OPPOSITE OF EXPANSION.

Factorising is the reverse of expanding. The general method used to factorise terms with common factors is to find the factors common to all the terms in the expression and bring it outside the bracket

Let us take another example

Factorise 3x²y + 12xy²z.

The highest common factor of 3 and 12 is 3. Also notice that x and y are common variables of both expressions. Therefore, the highest common factor of the expression above is 3xy.Write 3xy in front of a bracket. Divide 3x²y + 12xy²z by 3xy and write the remainder inside the bracket.

⇒ 3x²y + 12xy²z =3xy(x + 4yz)

Example 1

Factorize each expression

1) 4a + 6x                                         2) ax + bx

Solution

1) 4a + 6x

The highest common factor H.C.F of the term is 2

= 2×2a + 2 x               Enclose expression in a bracket

= (2×2a + 2 x 3x )

Place the common factor outside the bracket.

= 2(2a + 3x )

∴ 4a + 6x = 2(2a + 3x )

2) ax + bx                   Common factor is x

= ax + bx     

 Enclose expression in a bracket              

= (ax + bx)           

   Place the common factor outside the bracket.

 = x(a+b)

∴ ax + bx =x(a + b)

Example 2

Factorise each expression.

a) 6t + 9m                    b) 6my + 4py

c) 5k² – 25k                    d) 10a²b – 15ab²

Solution

a) 6t + 9m   

The highest common factor. (H.C.F) of the terms is 3

= 2×3t + 3×3m           Enclose expression in a bracket.

= (2×3t + 3×3m)

Place the common factor outside the bracket.

= 3(2t + 3m)

∴ 6t + 9m = 3(2t + 3m)

b) 6my + 4py

The common factor of the terms is  2y

= 2×3xm×y + 2×2p×y   

 Enclose expression in a bracket.                                                     = (2×3xm×y + 2×2p×y)       

 Place the common factor outside the bracket.

= 2y(3m + 2p)

∴ + 4py = 2y(3m + 2p)

c) 5k² – 25k  The H.C.F of the terms is 5k

= 5×k×k –  5×5×k 

Enclose expression in a bracket

=(5×k×k – 5×5×k)

Place the common factor outside the bracket.

=5k (k – 5k)

Therefore 5k² – 25k = 5k (k – 5k)

d) 10a²b –15ab²            The common factor of the terms is 5ab

= 2×5 a×a×b  –  3×5×a×b×b

Enclose expression in bracket     

= (2×5 a×a×b – 3×5×a×b×b)

 Place the common factor outside the bracket

= 5ab (2 – 3b)

Therefore 10a² –15ab²= 5ab (2 –3b)