## Monday, April 15, 2013

### Factorization Made Easy - Part I

FACTORIZATION: AN INTRODUCTION -Part I

When a number is written as a product of its factors, it is said factorization of the number. Let us see some example:

36=3 × 12

We can also write 36 as a product of its different factors, such as

36 = 4 × 9 or 36 = 18 × 2 or 36 = 2×2 × 3 ×3

In this example, numbers, 2, 3, 4, 6, 9, 12 and 18 are all factors of number 36. Number 2 and 3 are called the prime factor of 36

Therefore, 36 = 2 × 2 × 3 × 3 is called prime factorization of 36.

So, prime factors are those factors which cannot be further expressed as a product of factors. It is also called Irreducible factors.

Remarkable fact: The number 36 can be written as 1 × 36 also. So, 1 and 36 are also factors of the numbers 36.

For any number, 1 and the number itself are factors of the number. But, unless specifically asked to mention, we never mention these as factors.

1. FACTORIZATION OF ALGEBRAIC EXPRESSIONS:

Let us understand what is factorization of Algebraic Expression? It means writing it as a product of prime or irreducible factors. These factors may be algebraic Expression, constant or variable. Let us consider on an algebraic expression.

As we know that an algebraic expression consists of terms and terms are product of factors.
For an example: 7ab + 4c

In this expression, the term 7ab is obtained by multiplying 7, a, b. hence 7, a and b is factors of 7ab.
We can write it, 7ab = 7 × a × b

Similarly, in other expression, 24ab + 6c, the term 24ab can be written as the product of its factors in the following three different ways.

a. 24ab = 3 × 8 × a × b

b. 24ab = 2 × 2 × 3 × 3 × a × b

c. 24ab = 2 × 2 × 3 × 3 × ab

The factorization in (b) is the factorization of the term 24ab into irreducible factors. It means the term 24ab cannot be further expressed as a product of more irreducible factors. This factorization is called prime factorization or the irreducible factorization or irreducible form of the expression 24ab.

The factorization in (a) is not an irreducible of expression 24ab, since 8 can be factorized further. Similarly, (c) is not an irreducible form of expression 24ab, since ab can be factorized further and written as a × b.

Remarkable fact:- 1 is a factor of every term of an algebraic expression, for instant, the term 7ab can written as 1 × 7 × a × b.

But, as in factorization of numbers, we never write 1 as a factor of the terms of an algebraic expression unless we required it for some reason.

Let us consider on one more expression:- 6p (2q + 5r). This expression can be written as the product of its irreducible factors as:-

6p (2q + 5r) = 2 × 3 × p × (2q + 5r)

Some expression like 9xy, 2x (4y +7), 15abc (2a + b)(3a – 4b) are easy to factorize. But all algebraic expressions are not easy to factorize.

Let us learn how to factorize algebraic expressions systematically:

Factorization by finding common factors:-

Let us consider the expression 4ab + 6b.

First of all, we need to write the terms of a given expression as product of their irreducible factors.

First step:- Here, the irreducible factors of the terms 4ab and 6b are

4ab = 2 × 2 × a × b

6b = 2 × 3 × b

Second step: Search common factor:-

The common factor to both terms is 2 and b, so their HCF is 2b.

Now, we can write

4ab + 6b = (2 × 2 × a × b) + (2 × 3 × b)

= (2a × 2b) + (2b × 3)

= 2b (2a + 3)

[Using distributing property (2b × 2a) + (2b × 3) = 2b × (2a +3)]

So, by finding the common factor 2b we have factorize the expression 4ab + 6b into irreducible factors 2b and (2a + 3)

2. Factorization by Regrouping:

Sometime in algebraic expressing, all the terms of a given expression do not have common factors. In that case, we make two groups of terms which now have a common factor and then factorize the expression.

For example, let us consider on this expression: 15ab + 6a + 10b +4

All the four terms have no common factors. However, first and second terms have 3a as a common factor, and the third and fourth terms have 2 as a common factor.

Therefore, we combine first and second terms into one group, and third & fourth terms into another group.

15ab + 6a + 10b + 4 = [(3a + 5b) + (3a × 2)] + [(2 × 5b) + (2 × 2)]

= [3a (5b + 2)] + [2(5b + 2)]

Now, the expression consists of two terms which have the expression (5b + 2) common. Using the distributive property, we have,

15ab + 6a + 10b + 4 = (5b + 2) (3a + 2)

Let see another example:-

12ax –  4ab + 18bx – 6b2

=  [(4a × 3x) + (4a ×  – b)] + [(6b × 3x) + (6b ×  – b)]

= [4a(3x – b)] + [6b(3x – b)]

=(4a + 6b) (3x –b)
=2(2a + 3b) (3x – b)

Remarkable fact:- Though the regrouping of the terms in an expression can be done in different ways, all regrouping may not lead to a factorization. We need to learn how to regroup the terms in different way by trial and error for getting an expression that can actually be factorized.

Let see an example:-

18 (a + b)2 – 6(a +b)

As it is obvious,  6 (a + b) is the highest common factor of the terms of the given expression, then

18 (a + b)2 – 6(a +b) = [6(a + b) × 3 (a + b)] – [6(a + b) × 1]

= 6(a +b) [3(a +b) – 1]

Remarkable fact:- In the above example, we need to show 1 as a factor since the entire term 6(a + b) is the highest common factor.

Continue.. Part II