ML Aggarwal Solution Class 9 Chapter 4 Factorisation Exercise 4.1

 Exercise 4.1

Question 1

Sol :

(i) 8xy3+12x2y2

⇒4xy2(2y+3x) 

∴H.C.F of 8xy3 and 12x2y2 is 4xy2


(ii) 15ax3-9ax2

∴H.C.F of 15ax3 and 9ax2 is 3ax2

⇒3ax2(5x-3)


Question 2

Sol :

(i) 21py2-56py

∴H.C.F of 21py2 and 56py is 7py

⇒7py(3y-8)


(ii) 4x3-6x2

∴H.C.F of 4x3 and 6x2 is 2x2

⇒2x2(2x-3)


Question 3

(i) 2πr2-4πr

H.C.F of 2πr2 and 4πr is 2πr

∴2πr(r-2)


(ii) 18m+16n

H.C.F of 18m and 16n is 2

⇒2(9m+8n)


Question 4

(i) 25abc2-15a2b2c

∴H.C.F of 25abc2 and 15a2b2c is 5abc

∴⇒5abc(5c-3ab)


(ii) 28p2q2r-42pq2r2

∴H.C.F of 28p2q2r and 42pq2r2 is 14pq2r

∴⇒14pq2r(2p-3r)


Question 5

(i) 8x3-6x2+10x

∴H.C.F of 8x3 ,6x2 ,10x is 2x

∴⇒2x(4x2-3x+5)


(ii) 14mn+22m-62p

H.C.F of 14mn , 22m and 62p are 2

∴⇒2(7mn+11m-31p)


Question 6

(i) 18p2q2-24pq2+30p2q

H.C.F of 18p2q2,24pq2 and 30p2q is 6pq

⇒6pq(3pq-4q+5p)


(ii) 27a3b3-18a2b3+75a3b2

H.C.F of 27a3b3 ,18a2b2 and 75aa3b2 is 3a2b2

⇒3a2b2(9a-6b+25a)


Question 7

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

H.C.F of 15a(2p-3q) and 10b(2p-3q) is 5(2p-3q)

⇒5(2p-3q)[3a-2b]


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

H.C.F of 3a(x2+y2) and 6b(x2+y2) is 3(x2+y2)

⇒3(x2+y2)(a+2b)


Question 8

(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]


(ii) 14(a-3b)2-21p(a-3b)

H.C.F of 17(a-3b)3 and 21p(a-3b) is 7(a-3b)

⇒7(a-3b)[2(a-3b)2-3p]


Question 9

(i) 10a(2p+q)3-15b(2p+q)2+35(2p+q)

H.C.F is 5(2p+q)

⇒5(2p+q)[2a(2p+q)2-3b(2p+q)+7]


(ii) x(x2+y2-z2)+y(-x2-y2+z2)-z(x2+y2-z2)

H.C.F is x2+y2-z2

⇒(x2+y2-z2)[x-y-z]

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