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) 15ax3−9ax2
H.C.F of expression 15ax3 and 9ax2 is 3ax2
15ax3−9ax2=3ax2(5x−3)
Solution 2
Sol :
(i) 21py2−56py
H⋅C⋅F of 21py2 and 56py is 7py
∴21py2−56py⇒7py(3y−8)
(ii) 4x3−6x2
H.C.F of 4x3 and 6x2 is 2x2
∴4x3−6x2⇒2x2(2x−3)
Solution 3
Sol :
(i) 25abc2−15a2b2c
H.C.F of 25abc2 and 15a2b2c is 5abc
∴25abc2−15a2b2c⇒5abc(5c−3ab)
(ii)
x2yz+xy2z+xyz2
H.C.F of x2yz, xy2z and xyz2 is xyz
∴xyz(x+y+z)
Solution 4
Sol :
(i) 8x3−6x2+10x
H.C.F of 8x3,6x2 and 10x is 2x
⇒2x(4x2−3x+5)
(ii) 14mn+22 m−6p
H.C.F of 14mn,22 m and 62P is 2
⇒2(7mn+11n−31P)
Solution 5
Sol :
(i) 18p2q2−24pq2+30p2q
H⋅C⋅F of 18p2q2,24pa2 and 30p2q is 6pq
⇒6pq(3pq−4q+5p)
(ii) 27a3b3−18a2b3+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(2p−3q)(3a−2b)
(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(a−3b)3−21p(a−3b)
H.C.F of 14(a−3b)3 and 21p(a-3 b) is F(a-3 b)
→7(a−3b)[2(a−3b)2−3p]
Solution 8
Sol :
(i) 10a(2p+q)3−15b(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)(2a−3b+7)
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