## Book: NCERT - Mathematics

### Chapter: 14. Factorisation

#### Subject: Maths - Class 8th

##### Q. No. 2 of Exercises 14.1

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##### Factorise the following expressions.(i) (ii) (iii) (iv) (v) (vi) (vii)(viii)(ix) (x)

(i) Factor of 7x = 7 × x

Factor of 42 = 2× 3 × 7

Common factors is 7

7x-42 = (7× x)-( 2× 3 × 7) = 7(x-6)

(ii) Factor of 6p = 2 ×3 × p

Factor of -12q = -1× 2× 2 × 3 × q

Common factors is 2, 3 = 2× 3 = 6

6p-12q = (2 × 3 × p)-( 2× 2 × 3 × q) = 6(p-2q)

(iii) Factor of 7a2 = 7 × a × a

Factor of -12q = 2 × 7× a

Common factors is 7, a = 7 × a = 7a

7a2+14a = (7 × a × a)-( 2× 7 × a) = 7a(a-2)

(iv) Factor of -16z = - 2 × 2 × 2 × 2 × z

Factor of 20z3 = 2 × 2× 5 × z × z × z

Common factors is 2, 2, z = 2 × 2 × z = 4z

-16z+20z3 = -4z(z-5z2)

(v) Factor of 20l3m = 2 × 2 × 5 × l × l × l × m

Factor of 30alm = 2 × 3 × 5 × a × l × m

Common factors is 2, 5, l, m = 2 × 5 × l × m = 10lm

20l3m+30alm = 10lm(2l2-3a)

(vi) Factor of 5x2y = 5 × x × x × y

Factor of -15xy2 = -3 × 5 × x × y × y

Common factors is 5, x, y = 5 × x × y = 5xy

5x2y-15xy2= 5xy(x-3y)

(vii) Factor of 10a2 = 2 × 5 × a × a

Factor of -15b2 = -3 × 5 × b × b

Factor of 20c2 = 2 × 2 × 5 × c × c

Common factors is 5

10a2-15b2+20c2 = 5(2a2-3b2+c2)

(viii) Factor of -4a2 = -2 × 2 × a × a

Factor of 4ab = 2 × 2 × a × b

Factor of 4ca = 2 × 2 × c × a

Common factors is 2, 2, a

-4a2+4ab-4ca = 4a(-a+b-c)

(ix) Factor of x2yz = x × x × y × z

Factor of xy2z = x × y × y × z

Factor of xyz2 = x × y × z × z

Common factors is x, y, z

X2yz + xy2z + xyz2 = xyz(z+y+z)

(x) Factor of ax2y = a × x × x × y

Factor of bxy2 = b × x × y × y

Factor of cxyz = c × x × y × z

Common factors is x, y

a x2y + bxy2 + cxyz = xy(ax+by+cz)

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