Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

Factorisation : Exercise 14.1 (Mathematics NCERT Class 8th)


Q.1 Find the common factors of the given terms
(i) 12x,36
(ii)
2y,22xy
(iii)
14pq,28{p^2}{q^2}
(iv) 2x,3{x^2},4
(v)
6abc,24a{b^2},12{a^2}b
(vi)
16{x^3}, - 4{x^2},32x
(vii) 10pq,20qr,30rp
(viii)
3{x^2}{y^3},10{x^3}{y^2},6{x^2}{y^2}z
Sol. (i) 12x,36
The factors of 12x = 2 \times 2 \times 3 \times x
The factors of 36 = 2 \times 2 \times 3 \times 3
Therefore, the common factors are {\rm{2, 2 and 3 = }}2 \times 2 \times 3 = 12

(ii) 2y,22xy
The factors of 2y = 2 \times y
The factors of 22xy = 2 \times 11 \times x \times y
Therefore, the common factors are {\rm{2 and }}y{\rm{ = }}2 \times y = 2y

(iii) 14pq,28{p^2}{q^2}
The factors of 14pq = 2 \times 7 \times p \times q
The factors of 28{p^2}{q^2} = 2 \times 2 \times 7 \times p \times p \times q \times q
Therefore, the common factors are .{\rm{2, 7, }}p{\rm{ and }}q{\rm{ = }}2 \times 7 \times p \times q = 14pq.

(iv) 2x,3{x^2},4
The factors of 2x = 2 \times x \times 1
The factors of 3{x^2} = 3 \times x \times x \times 1
The factors of 4 = 2 \times 2 \times 1
Therefore, the common factor is 1.

(v) 6abc,24a{b^2},12{a^2}b
The factors of 6abc = 2 \times 3 \times a \times b \times c
The factors of 24a{b^2} = 2 \times 2 \times 2 \times 3 \times a \times b \times b
The factors of 124{a^2}b = 2 \times 2 \times 3 \times a \times a \times b
Therefore, the common factors are {\rm{2, 3, }}a{\rm{ and }}b{\rm{ = }}2 \times 3 \times a \times b = 6ab

(vi) 16{x^3}, - 4{x^2},32x
The factors of 16{x^3} = 2 \times 2 \times 2 \times 2 \times x \times x \times x
The factors of  - 4{x^2} = ( - 1) \times 2 \times 2 \times x \times x
The factors of 32x = 2 \times 2 \times 2 \times 2 \times 2 \times x
Therefore, the common factors are {\rm{2, 2 and }}x{\rm{ = }}2 \times 2 \times x = 4x

(vii) 10pq,20qr,30rp
The factors of 10pq = 2 \times 5 \times p \times q
The factors of 20qr = 2 \times 2 \times 5 \times q \times r
The factors of 30rp = 2 \times 3 \times 5 \times r \times p
Therefore, the common factors are {\rm{2 and 5 = }}2 \times 5 = 10

(viii) 3{x^2}{y^3},10{x^3}{y^2},6{x^2}{y^2}z
The factors of 3{x^2}{y^3} = 3 \times x \times x \times y \times y \times y
The factors of 10{x^3}{y^2} = 2 \times 5 \times x \times x \times x \times y \times y
The factors of 6{x^2}{y^2}z = 2 \times 3 \times x \times x \times y \times y \times z
Therefore, the common factors are x,{\rm{ }}x,{\rm{ }}y{\rm{ and }}y{\rm{ = }}x \times x \times y \times y = {x^2}{y^2}

Q.2 Factorise the following expressions.
(i) 7x - 42
(ii)
6p - 12q
(iii)
7{a^2} + 14a
(iv)  - 16z + 20{z^3}
(v)
20{l^2}m + 30alm
(vi)
5{x^2}y - 15x{y^2}
(vii) 10{a^2} - 15{b^2} + 20{c^2}
(viii)
 - 4{a^2} + 4ab - 4ca
(ix)
{x^2}yz + x{y^2}z + xy{z^2}
(x) a{x^2}y + bx{y^2} + cxyz
Sol. (i) 7x - 42
7x - 42=7 \times x - 2 \times 3 \times 7
= 7(x - 2 \times 3)
= 7(x - 6)

(ii)6p - 12q
6p - 12q = 2 \times 3 \times p - 2 \times 2 \times 3 \times q
= 2 \times 3(p - 2q)
= 6(p - 2q)

(iii) 7{a^2} + 14a
7{a^2} + 14a = 7 \times a \times a + 2 \times 7 \times a
= 7 \times a(a + 2)
= 7a(a + 2)

(iv)  - 16z + 20{z^3}
 - 16z + 20{z^3} = ( - 1) \times 2 \times 2 \times 2 \times 2 \times z + 2 \times 2 \times 5 \times z \times z \times z
= 2 \times 2 \times z( - 2 \times 2 + 5 \times z \times z)
= 4z( - 4 + 5{z^2})

(v) 20{l^2}m + 30alm
20{l^2}m + 30alm = 2 \times 2 \times 5 \times l \times l \times m + 2 \times 3 \times 5 \times a \times l \times m
= 2 \times 5 \times l \times m(2 \times l + 3 \times a)
= 10lm(2l + 3a)

(vi) 5{x^2}y - 15x{y^2}
5{x^2}y - 15x{y^2} = 5 \times x \times x \times y + 3 \times 5 \times x \times y \times y
= 5 \times x \times y(x - 3 \times y)
= 5xy(x - 3y)

(vii) 10{a^2} - 15{b^2} + 20{c^2}
10{a^2} - 15{b^2} + 20{c^2} = 2 \times 5 \times a \times a - 3 \times 5 \times b \times b + 2 \times 2 \times 5 \times c \times c
= 5(2 \times a \times a - 3 \times b \times b + 2 \times 2 \times c \times c)
= 5(2{a^2} - 3{b^2} + 4{c^2})

(viii)  - 4{a^2} + 4ab - 4ca
 - 4{a^2} + 4ab - 4ca = ( - 1) \times 2 \times 2 \times a \times a + 2 \times 2 \times a \times a - 2 \times 2 \times c \times a
= 2 \times 2 \times a( - a + b - c)
= 4a( - a + b + c)

(ix) {x^2}yz + x{y^2}z + xy{z^2}
{x^2}yz + x{y^2}z + xy{z^2} = x \times x \times y \times z + x \times y \times y \times z + x \times y \times z \times z
= x \times y \times z(x + y + z)
= xyz(x + y + z)

(x) a{x^2}y + bx{y^2} + cxyz
a{x^2}y + bx{y^2} + cxyz = a \times x \times x \times y + b \times x \times y \times y + c \times x \times y \times z
= x \times y(a \times x + b \times y + c \times 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
Sol. (i) {x^2} + xy + 8x + 8y
{x^2} + xy + 8x + 8y = x(x + y) + 8(x + y)
= (x + y)(x + 8)

(ii) 15xy - 6x + 5y - 2
15xy - 6x + 5y - 2 = 3x(5y - 2) + 1(5y - 2)
= (5y - 2)(3x + 1)

(iii)ax + bx - ay - by
ax + bx - ay - by = x(a + b) - y(a + b)
= (a + b)(x - y)

(iv) 15pq + 15 + 9q + 25p
15pq + 15 + 9q + 25p = 5p(3q + 5) + 3(3q + 5)
= (3q + 5)(5p + 3)

(v) z - 7 + 7xy - xyz
z - 7 + 7xy - xyz = 7xy - 7 - xyz + z
= 7(xy - 1) - z(xy - 1)
= (xy - 1)(7 - z)



1 Comment

Leave a Reply

Contact Us

Call us: 8287971571,0261-4890014

Or, Fill out the form & get a call back.!