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7. Factorisation
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Q2 of 137 Page 99

Factories:

i. 16a2 – 24ab


ii. 15ab2 – 20a2b


iii. 12x2y3 – 21x3y2

(i) Let’s take HCF of 16a2 – 24ab


Taking 8a as common from the whole, we get,


16a2 – 24ab = 8a(2a – 3b).


(ii) 15ab2 – 20a2b,


Taking 5ab as common from the whole, we get,


15ab2 – 20a2b = 5ab(3b – 4a)


(iii) 12x2y3 – 21x3y2,


Taking 3x2y2 as common from the whole, we get,


12x2y3 – 21x3y2 = 3x2y2(4y – 7x)


More from this chapter

All 137 →
1

Factories:

(i) 12x + 15


(ii) 14m – 21


(iii) 9n – 12n2

3

Factories:

(i) 24x3 – 36x2y


(ii) 10x3 – 15x2


(iii) 36x3y – 60x2y3z

4

Factories:

i. 9x3 – 6x2 + 12x


ii. 8x3 – 72xy + 12x


iii. 18a3b3-27a2b3+36a3b2

5

Factories:

i. 14x3 + 21x4y – 28x2y2


ii. - 5 – 10t + 20t2

Questions · 137
7. Factorisation
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