Page 57 - SM inner class 8.cdr

2
2
2
Solution: The HCF of 35a , 21a b and 14a b is Common factor from 2nd group = 3
7a 2 We can write the above term like as:
2
2
2
2
So, 35a - 21a b + 14a b = 2y x + 1) + 3 x + 1)
(
(
2
2
= 7a ( 5 - 3b + 2b ) Now there are two terms i.e 2y x( + 1) and
3(x + 1), take com mon bi no mial fac tor (x + 1 )
2. Factorization by Grouping
So, (x + 1 ) ( y + 3 ) are the factors of
2
In factorization all the terms in a given expres- 2xy + 3 + 2y + 3x.
sion do not have a common factor; but the Example 5. Factorize 6xy - 4y + 6 - 9x.
terms can be rearranged in such a way that all
Solution: 6xy - 4y + 6 - 9x
the terms in each group have a common factor.
Rearranging the expression, as 6xy - 9x -
In factorization by regrouping, we should 4y + 6
remember that any regrouping of the terms in = 6xy - 9x - 4y + 6
the given expression may not lead to factorizat-
[these are the two groups]
ion. We must observe the expression and come
out with the desired regrouping by trial and Common factor from 1st group = 3x
error. Common factor from 2nd group = -2
= 3x ( 2y - 3) -2 2y - 3)
(
Steps of Factorization by Grouping :
Now there are two terms, take common
1. Rearranging the expression so as to form binomial factor (2y - ) 3
groups. So, (2y - 3 ) (3x - ) 2 are the factors of
6xy - 4y + 6 - 9x.
2. Find the common factors from each group.
Example 6. Factorize: 15pq + 15 + 9q + 25p
Example 4. 2xy + 3 + 2y + 3x
Solution: 2xy + 3 + 2y + 3x Solution: 15pq + 15 + 9q + 25p
=15pq + 9q + 15 + 25p
Rearranging the expression, as
(
2xy + 2y + 3x + 3 = 3q ( 5p + 3) + 5 3 + 5p)
(
(
q
= 3 5p + 3) + 5 5p + 3)
= 2xy + 2y + 3x + 3
5
= (5p + ) (3q + )
3
[these are the two groups]
Common factor from 1st group = 2y
E xercise 7.1
1. Find the common factors of the given terms in each:
2
2
2
(i) 8 24, (ii) 3 21a, ab (iii) 7xy, 35x y 3 (iv) 4m , 6m , 8m 3
2
2
2
(v) 15 20p, qr, 25rp (vi) 4x , 6xy, 8y x (vii) 12x y, 18xy 2
2. Factor ise the follow ing expres sions by taking out a common factor :
2
2
2
2
2
(i) 6x - 25xy (ii) 9a - 6ax (iii) 7p + 49pq (iv) 36a b - 60a bc
2
2
2
(v) 3a bc + 6ab c + 9abc 2 (vi) 4p + 5pq - 6pq 2 (vii) ut at+ 2
3. Factor ise the follow ing by regroup ing :
3
(i) 8ax - 8xy + 4by - 4ab (ii) x + x 2 2 + x 5 + 10
2
2
2
3
+
(iii) m - mn 4 m 4 n (iv) a - a b - ab b 3
-
+
2
(v) p q pr- 2 - pq r 2
+
3. Factorization using Identities There are some identities that can be used to
make the factorization much easier.
57
Mathematics In Focus - 8