ML AGGARWAL CLASS 8 CHAPTER 11 Factorisation Exercise 11.5
Exercise 11.5
Solution 1
Sol :
(i) (35x+28)÷(5x+4)
35x+285x+4
⇒7(5x+4)5x+4
⇒7
(ii) 7p2q2(9r−21)÷63pq(r−3)
⇒ 7p2q2(9r−27)63pq(r−3)
⇒ 7⋅p2q2⋅q(r−3)63pq(x−3)
⇒ pq
Solution 2
Sol :
(i) 6(2x+7)(5x−3)÷3(5x−3)
⇒6(2x+7)(5x−3)3(5x−3)
⇒2(2 x+7)
⇒4x+14
(ii) 33pq(p+3)(2q−5)÷11p(2q−5)
⇒38pq(p+3)(2q−8)11−pq(2q−5)
⇒3(p+3)
⇒3p+9
Solution 3
Sol :
(i) (7x3−63x)÷7(x−3)
⇒ 7x3−6x7(x−3)
⇒7x(x2−9)7(x−3)
⇒x(x2−32)x−3
⇒x(x+3)(x−3)(x−3)
⇒x(x+3)
⇒x2+3x
(ii) (3p2+17p+10)÷(p+5)
⇒3p2+17p+10p+5
⇒3p2+15p+2p+10p+5
⇒3P(p+5)+2(p+5)p+5
⇒(p+5)(3p+2)(p+5)
⇒ 3p+2
(iii) 10xy(14y2+43y−21)÷5x(7y−3)
⇒ 10y(14y2+43y−21)5x(7y−3)
⇒2y(14y2−6y+49y−21)7y−3
⇒2y(2y(7y−3)+7(7y−3))7y−3
⇒2y(7y−3)(2y+7)(7y−3)
⇒2y(2 y+7)
(iv) 12pqr(6p2−13pq+6q2)÷6pq(2p−3q)
⇒ 12pqr(6p2−13pq+6q2)6pq(2p−3q)
⇒ 2r(6p2−9pq−4pq+6q2)2p−3q
⇒ 2r(3p(2p−3q)−2q(2p−3q))2p−3q
⇒ 2r(2p−3q)(3p−2q)(2p−3q)
⇒2(3 p-2 q)
Comments
Post a Comment