ML AGGARWAL CLASS 8 CHAPTER 11 Factorisation Exercise 11.1

 Exercise 11.1

Solution 1

Sol :

(i) 8xy3+12x2y2

H.C.F of 8xy3 and 12x2y2 is 4xy2

Divide each expression by 4xy2 and keep 4xy2
outside the bracket.

4xy2(2y+3x)

8xy3+12x2y2=4xy2(2y+3x)


(ii) 15ax39ax2

H.C.F of expression 15ax3 and 9ax2 is 3ax2

15ax39ax2=3ax2(5x3)

Solution 2

Sol :
(i) 21py256py

HCF of 21py2 and 56py is 7py

21py256py7py(3y8)


(ii) 4x36x2

H.C.F of 4x3 and 6x2 is 2x2

4x36x22x2(2x3)

Solution 3

Sol :

(i) 25abc215a2b2c

H.C.F of 25abc2 and 15a2b2c is 5abc

25abc215a2b2c5abc(5c3ab)


(ii)
 x2yz+xy2z+xyz2

H.C.F of x2yz, xy2z and xyz2 is xyz

xyz(x+y+z)

Solution 4

Sol :
(i) 8x36x2+10x

H.C.F of 8x3,6x2 and 10x is 2x

2x(4x23x+5)


(ii) 14mn+22 m6p

H.C.F of 14mn,22 m and 62P is 2

2(7mn+11n31P)


Solution 5

Sol :
(i) 18p2q224pq2+30p2q

HCF of 18p2q2,24pa2 and 30p2q is 6pq

6pq(3pq4q+5p)


(ii) 27a3b318a2b3+75a3b2

H. C .F of 27a3b3,18a2b2 and 7sa3b2 is 3a2b2

$\Rightarrow 3 a^{2} b^{2}(9 a b-6 b+25 a)

Solution 6

Sol :
(i) 15a(2p-3q)-10b(2p-3q)

HCFof 15 a(2 p-3 q) and 10b(2 p-3 q) is 5(2 p-3 q)

5(2p3q)(3a2b)


(ii) 3a(x2+y2)+6b(x2+y2)

HCF of 3a(x2+y2) and 6b(x2+y2) is 3(x2+y2)

∴⇒3(x2+y2)(a+2b)

Solution 7

Sol :
(i) 6(x+2y)3+8(x+2y)2

H.C.F of 6(x+2y)3 and 8(x+2y)2 is 2(x+2y)2

2(x+2y)2(3(x+2y)+4)

2(x+2y)2(3x+6y+4)



(ii)14(a3b)321p(a3b)

H.C.F of 14(a3b)3 and 21p(a-3 b) is F(a-3 b)

7(a3b)[2(a3b)23p]

Solution 8

Sol :
(i) 10a(2p+q)315b(2p+q)2+35(2p+q)

HCF of 10a(2p+q)3,15b(2p+q)2 and

35(2p+q) is 5(2p+q)

5(2p+q)(2a3b+7)

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