Class VIII NCERT (CBSE) Mathematics | Factorisation (Exercise 14.1)

Class 8, NCERT (CBSE) Mathematics

Chapter 14, Mensuration

Textbook Exercise 14.1 Solution

 (Page 220, 221)

Q.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

Ans:

 (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

Q. 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

Ans:

 (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) + b − c]

= 4a (−a + b − c)

(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)

Q.3: Factorise:

(i) x² + xy + 8x + 8y

(ii) 15xy − 6x + 5y − 2

(iii) ax + bx − ay − by

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

(v) z − 7 + 7xy − xyz

Ans:

(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 + bx − ay − by = a × x + b × x − a × y − b × y

= x (a + b) − y (a + b)

= (a + b) (x − y)

(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 + 7xy − xyz = z − x × y × z − 7 + 7 × x × y

= z (1 − xy) − 7 (1 − xy)

= (1 − xy) (z − 7)