NCERT solution class 8 chapter 14 Factorisation

Exercise 14.1   Exercise 14.2   Exercise 14.3   Exercise 14.4

Exercise 14.1

Question 1 

Find the common factors of the terms

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28p2q2

(iv) 2x, 3x2, 4

(v) 6abc, 24ab2, 12a2b

(vi) 16x3, −4x2, 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x2y3, 10x3y2, 6x2y2z

Sol :

(i) 12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

The common factors are 2, 2, 3.

And, 2 × 2 × 3 = 12

(ii) 2y = 2 × y

22xy = 2 × 11 × x × y

The common factors are 2, y.

And, 2 × y = 2y

(iii) 14pq = 2 × 7 × p × q

28p2q2 = 2 × 2 × 7 × p × p × q × q

The common factors are 2, 7, p, q.

And, 2 × 7 × p × q = 14pq

(iv) 2x = 2 × x

3x2 = 3 × x × x

4 = 2 × 2

The common factor is 1.

(v) 6abc = 2 × 3 × a × b × c

24ab2 = 2 × 2 × 2 × 3 × a × b × b

12a2b = 2 × 2 × 3 × a × a × b

The common factors are 2, 3, a, b.

And, 2 × 3 × a × b = 6ab

(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x

−4x2 = −1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

The common factors are 2, 2, x.

And, 2 × 2 × x = 4x

(vii) 10pq = 2 × 5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

The common factors are 2, 5.

And, 2 × 5 = 10

(viii) 3x2y3 = 3 × x × x × y × y × y

10x3y2 = 2 × 5 × x × x × x × y × y

6x2y2z = 2 × 3 × x × x × y × y × z

The common factors are x, x, y, y.

And,

x × x × y × y = x2y2

 

 

Question 2

Factorise the following expressions

(i) 7x − 42

(ii) 6p − 12q

(iii) 7a2 + 14a

(iv) −16z + 20z3

(v) 20l2m + 30 alm

(vi) 5x2y − 15xy2

(vii) 10a2 − 15b2 + 20c2

(viii) −4a2 + 4ab − 4 ca

(ix) x2yz + xy2z + xyz2

(x) ax2y + bxy2 + cxyz

Sol :

(i) 7x = 7 × x

42 = 2 × 3 × 7

The common factor is 7.

∴ 7x − 42 = (7 × x) − (2 × 3 × 7) = 7 (x − 6)

(ii) 6p = 2 × 3 × p

12q = 2 × 2 × 3 × q

The common factors are 2 and 3.

∴ 6p − 12q = (2 × 3 × p) − (2 × 2 × 3 × q)

= 2 × 3 [p − (2 × q)]

= 6 (p − 2q)

(iii) 7a2 = 7 × a × a

14a = 2 × 7 × a

The common factors are 7 and a.

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

= 7 × a [a + 2] = 7a (a + 2)

(iv) 16z = 2 × 2 × 2 × 2 × z

20z3 = 2 × 2 × 5 × z × z × z

The common factors are 2, 2, and z.

∴ −16z + 20z3 = − (2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)

= (2 × 2 × z) [− (2 × 2) + (5 × z × z)]

= 4z (− 4 + 5z2)

(v) 20l2m = 2 × 2 × 5 × l × l × m

30alm = 2 × 3 × 5 × a × l × m

The common factors are 2, 5, l, and m.

∴ 20l2m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)

= (2 × 5 × l × m) [(2 × l) + (3 × a)]

= 10lm (2l + 3a)

(vi) 5x2y = 5 × x × x × y

15xy2 = 3 × 5 × x × y × y

The common factors are 5, x, and y.

∴ 5x2y − 15xy2 = (5 × x × x × y) − (3 × 5 × x × y × y)

= 5 × x × y [x − (3 × y)]

= 5xy (x − 3y)

(vii) 10a2 = 2 × 5 × a × a

15b2 = 3 × 5 × b × b

20c2 = 2 × 2 × 5 × c × c

The common factor is 5.

10a2 − 15b2 + 20c2 = (2 × 5 × a × a) − (3 × 5 × b × b) + (2 × 2 × 5 × c × c)

= 5 [(2 × a × a) − (3 × b × b) + (2 × 2 × c × c)]

= 5 (2a2 − 3b2 + 4c2)

(viii) 4a2 = 2 × 2 × a × a

4ab = 2 × 2 × a × b

4ca = 2 × 2 × c × a

The common factors are 2, 2, and a.

∴ −4a2 + 4ab − 4ca = − (2 × 2 × a × a) + (2 × 2 × a × b) − (2 × 2 × c × a)

= 2 × 2 × a [− (a) + bc]

= 4a (−a + bc)

(ix) x2yz = x × x × y × z

xy2z = x × y × y × z

xyz2 = x × y × z × z

The common factors are x, y, and z.

x2yz + xy2z + xyz2 = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)

= x × y × z [x + y + z]

= xyz (x + y + z)

(x) ax2y = a × x × x × y

bxy2 = b × x × y × y

cxyz = c × x × y × z

The common factors are x and y.

ax2y + bxy2 + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)

= (x × y) [(a × x) + (b × y) + (c × z)]

= xy (ax + by + cz)

 

 

Question 3

Factorise

(i) x2 + xy + 8x + 8y

(ii) 15xy − 6x + 5y − 2

(iii) ax + bxayby

(iv) 15pq + 15 + 9q + 25p

(v) z − 7 + 7xyxyz

Sol :

(i) x2 + xy + 8x + 8y = x × x + x × y + 8 × x + 8 × y

= x (x + y) + 8 (x + y)

= (x + y) (x + 8)

(ii) 15xy − 6x + 5y − 2 = 3 × 5 × x × y − 3 × 2 × x + 5 × y − 2

= 3x (5y − 2) + 1 (5y − 2)

= (5y − 2) (3x + 1)
​​​​​​​​​​​​​​​​​​​​​

(iii) ax + bxayby = a × x + b × xa × yb × y

= x (a + b) − y (a + b)

= (a + b) (xy)

(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15

= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5

= 3q (5p + 3) + 5 (5p + 3)

= (5p + 3) (3q + 5)

(v) z − 7 + 7xyxyz = zx × y × z − 7 + 7 × x × y

= z (1 − xy) − 7 (1 − xy)

= (1 − xy) (z − 7)

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