Page 56 - SM inner class 8.cdr

+ 4 % 1 ¼ 3 ÷ ¾

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Chapter7 6 ½ –

Factorisation

(
Introduction It can be written as 5 ´ p ´ q + 3 ´ r)
r
In previous classes, we have learnt about Here 5, p (q + 3 ) are the irreducible factors.
(
)
factors. Let us recall how to factorize the Similarly: 3x y( + 5 z + 7) = 3 ´ x ´ y ( + 5)
natural number. ´ (z + ) Here 3, x , (y + 5), (z + 7) are irreducible
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We know, 15 = 3 ´ 5 or 1 15 factors.
´
So, we say that 1, 3, 5 and 15 are the factors of The factorization helps to write the algebraic
15. expression in simpler form and it also helps in
simplifying the algebraic expression.
Among the above factors, there are some prime
numbers. i.e. 3 ´ 5 There are four ways to factorized an algebraic
expression.
So 3 and 5 are prime factors of 15.
(i) Factorisation by taking out a common
And 15 = 5 ´ 3 is called prime factorization of
factor.
15.
(ii) Factorisation by using grouping:
Prime factors are irreducible factors. This
(iii) Factorisation by using Identifies.
means that they cannot be further divided into
(iv) Factorisation by splitting the middle
smaller factors.
term.
Factors of algebraic expressions: 1. Factorization by taking out a Common
When an algebraic expression can be written Factor
as the product of two or more expressions, then A systematic way of factorizing an expression
each of these expression is called a factor of the is the Factorization by Common Factor. It
given expression. consists of three steps:
Consider the following example : Step 1: Write each term of the expression
Lets do the Factorization of 13ab. as a product of irreducible factors.
Step 2: Look for and separate the common
(
13ab = 13 ab) ( Here,13 and ab are the factors)
factors.
= 13a(b) ( Here,13a and b are the factors)
Step 3: Combine the remaining factors in
= 13b(a) ( Here,13b and a are the factors) each term in accordance with the
= 13(a)(b) ( Here,13, a and b are the factors) distributive law.

Among the above factors 13, a, b are Example 1. Factorize : 8x + 8
irreducible factors. The phrase ‘irreducible’ is Solution: The HCF of 8x and 8 is 8.
used in the place of ‘prime’ in algebraic 8x + 8 = 8 (x + 1)
expressions. Thus, we say that 13 ´ a ´ b is the
2
Example 2. Factorize: 4x - 6xy + 12x
irreducible form of 13ab.
2
Solution: The HCF of 4x , 6xy and 12x is 2x.
Note : 13 ´ (ab ) or 13a b( ) or 13b a( ) are not in 4x - 6xy + 12x = 2x ( 2x -3y + 6)
2
irreducible form.
Example 3. Factorize:
2
2
2
2
Let us take the expression 5p q( + 3r) 35a - 21a b + 14a b

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Mathematics In Focus - 8