# 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) $z - 7 + 7xy - xyz$
$z - 7 + 7xy - xyz = 7xy - 7 - xyz + z$
= $7(xy - 1) - z(xy - 1)$
= $(xy - 1)(7 - z)$

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