**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, 28p

^{2}q^{2}(iv) 2x, 3x

^{2}, 4(v) 6abc, 24ab

^{2}, 12a^{2}b(vi) 16x

^{3}, −4x^{2}, 32x(vii) 10pq, 20qr, 30rp

(viii) 3x

^{2}y^{3}, 10x^{3}y^{2}, 6x^{2}y^{2}zAns:

(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

28p

^{2}q^{2}= 2 × 2 × 7 × p × p × q × qThe common factors are 2, 7, p, q.

And, 2 × 7 × p × q = 14pq

(iv) 2x = 2 × x

3x

^{2}= 3 × x × x4 = 2 × 2

The common factor is 1.

(v) 6abc = 2 × 3 × a × b × c

24ab

^{2}= 2 × 2 × 2 × 3 × a × b × b12a

^{2}b = 2 × 2 × 3 × a × a × bThe common factors are 2, 3, a, b.

And, 2 × 3 × a × b = 6ab

(vi) 16x

^{3}= 2 × 2 × 2 × 2 × x × x × x−4x

^{2}= −1 × 2 × 2 × x × x32x = 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) 3x

^{2}y^{3 }= 3 × x × x × y × y × y10x

^{3}y^{2}= 2 × 5 × x × x × x × y × y6x

^{2}y^{2}z = 2 × 3 × x × x × y × y × zThe common factors are x, x, y, y.

And,

x × x × y × y = x

^{2}y^{2}

Q. 2: Factorise the following expressions:

(i) 7x − 42

(ii) 6p − 12q

(iii) 7a

^{2}+ 14a(iv) −16z + 20z

^{3}(v) 20l

^{2}m + 30 alm(vi) 5x

^{2}y − 15xy^{2}(vii) 10a

^{2}− 15b^{2}+ 20c^{2}(viii) −4a

^{2}+ 4ab − 4 ca(ix) x

^{2}yz + xy^{2}z + xyz^{2}(x) ax

^{2}y + bxy^{2}+ cxyzAns:

(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) 7a

^{2}= 7 × a × a14a = 2 × 7 × a

The common factors are 7 and a.

∴ 7a

^{2}+ 14a = (7 × a × a) + (2 × 7 × a)= 7 × a [a + 2] = 7a (a + 2)

(iv) 16z = 2 × 2 × 2 × 2 × z

20z

^{3}= 2 × 2 × 5 × z × z × zThe common factors are 2, 2, and z.

∴ −16z + 20z

^{3}= − (2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)= (2 × 2 × z) [− (2 × 2) + (5 × z × z)]

= 4z (− 4 + 5z

^{2})(v) 20l

^{2}m = 2 × 2 × 5 × l × l × m30alm = 2 × 3 × 5 × a × l × m

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

∴ 20l

^{2}m + 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) 5x

^{2}y = 5 × x × x × y15xy

^{2}= 3 × 5 × x × y × yThe common factors are 5, x, and y.

∴ 5x

^{2}y − 15xy^{2}= (5 × x × x × y) − (3 × 5 × x × y × y)= 5 × x × y [x − (3 × y)]

= 5xy (x − 3y)

(vii) 10a

^{2}= 2 × 5 × a × a15b

^{2}= 3 × 5 × b × b20c

^{2}= 2 × 2 × 5 × c × cThe common factor is 5.

10a

^{2}− 15b^{2}+ 20c^{2}= (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 (2a

^{2}− 3b^{2}+ 4c^{2})(viii) 4a

^{2}= 2 × 2 × a × a4ab = 2 × 2 × a × b

4ca = 2 × 2 × c × a

The common factors are 2, 2, and a.

∴ −4a

^{2}+ 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) x

^{2}yz = x × x × y × zxy

^{2}z = x × y × y × zxyz

^{2}= x × y × z × zThe common factors are x, y, and z.

∴ x

^{2}yz + xy^{2}z + xyz^{2}= (x × x × y × z) + (x × y × y × z) + (x × y × z × z)= x × y × z [x + y + z]

= xyz (x + y + z)

(x) ax

^{2}y = a × x × x × ybxy

^{2}= b × x × y × ycxyz = c × x × y × z

The common factors are x and y.

ax

^{2}y + bxy^{2}+ 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) x

^{2}+ 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) x

^{2}+ 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)

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