# Factorisation : Exercise 14.3 (Mathematics NCERT Class 8th)

Q.1Carry out the following divisions.
(i) $28{x^4} \div 56x$
(ii)
$- 36{y^3} \div 9{y^2}$
(iii)
$66p{q^2}{r^3} \div 11q{r^2}$
(iv) $34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}$
(v)
$12{a^8}{b^8} \div ( - 6{a^6}{b^4})$
Sol. (i) $28{x^4} \div 56x$
$28{x^4} \div 56x = {{28{x^4}} \over {56x}} = {{28} \over {56}} \times {{{x^4}} \over x}$
$= {1 \over 2} \times {x^3}$

(ii) $- 36{y^3} \div 9{y^2}$
$- 36{y^3} \div 9{y^2} = {{ - 36{y^3}} \over {9{y^2}}} = {{ - 36} \over 9} \times {{{y^3}} \over {{y^2}}}$
$= - 4y$

(iii) $66p{q^2}{r^3} \div 11q{r^2}$
$66p{q^2}{r^3} \div 11q{r^2} = {{66p{q^2}{r^3}} \over {11q{r^2}}} = {{66} \over {11}} \times {{p{q^2}{r^3}} \over {q{r^2}}}$
$= 6pqr$

(iv) $34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}$
$34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3} = {{34{x^3}{y^3}{z^3}} \over {51x{y^2}{z^3}}} = {{34} \over {51}} \times {{{x^3}{y^3}{z^3}} \over {x{y^2}{z^3}}}$
$= {2 \over 3}{x^2}y$

(v) $12{a^8}{b^8} \div ( - 6{a^6}{b^4})$
$12{a^8}{b^8} \div ( - 6{a^6}{b^4}) = {{12{a^8}{b^8}} \over { - 6{a^6}{b^4}}} = {{12} \over { - 6}} \times {{{a^8}{b^8}} \over {{a^6}{b^4}}}$
$= - 2{a^2}{b^4}$

Q.2 Divide the given polynomial by the given monomial.
(i) $(5{x^2} - 6x) \div 3x$
(ii)
$(3{y^8} - 4{y^6} + 5{y^4}) \div {y^4}$
(iii) $8({x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}) \div 4{x^2}{y^2}{z^2}$
(iv)
$({x^3} + 2{x^2} + 3x) \div 2x$
(v) $({p^3}{q^6} - {p^6}{q^3}) \div {p^3}{q^3}$
Sol. (i) $(5{x^2} - 6x) \div 3x$
= $(5{x^2} - 6x) \div 3x = {{(5{x^2} - 6x)} \over {3x}}$
$= {{5{x^2}} \over {3x}} - {{6x} \over {3x}} = {5 \over 3}x - 2$
$= {1 \over 3}(5x - 6)$

(ii) $(3{y^8} - 4{y^6} + 5{y^4}) \div {y^4}$
= $(3{y^8} - 4{y^6} + 5{y^4}) \div {y^4} = {{3{y^8} - 4{y^6} + 5{y^4}} \over {{y^4}}}$
$= {{3{y^8}} \over {{y^4}}} - {{4{y^6}} \over {{y^4}}} + {{5{y^4}} \over {{y^4}}} = 3{y^4} - 4{y^2} + 5$

(iii) $8({x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}) \div 4{x^2}{y^2}{z^2}$
= $8({x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}) \div 4{x^2}{y^2}{z^2} = {{8({x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3})} \over {4{x^2}{y^2}{z^2}}}$
$= {{8{x^3}{y^2}z} \over {4{x^2}{y^2}{z^2}}} + {{8{x^2}{y^3}{z^2}} \over {4{x^2}{y^2}{z^2}}} + {{8{x^2}{y^2}{z^3}} \over {4{x^2}{y^2}{z^2}}}$
$= 2x + 2y + 2z = 2(x + y + z)$

(iv) $({x^3} + 2{x^2} + 3x) \div 2x$
= $({x^3} + 2{x^2} + 3x) \div 2x = {{{x^3} + 2{x^2} + 3x} \over {2x}}$
$= {{{x^3}} \over {2x}} + {{2{x^2}} \over {2x}} + {{3x} \over {2x}} = {1 \over 2}({x^2} + 2x + 3)$

(v) $({p^3}{q^6} - {p^6}{q^3}) \div {p^3}{q^3}$
= $({p^3}{q^6} - {p^6}{q^3}) \div {p^3}{q^3} = {{{p^3}{q^6} - {p^6}{q^3}} \over {{p^3}{q^3}}}$
= ${{{p^3}{q^6}} \over {{p^3}{q^3}}} - {{{p^6}{q^3}} \over {{p^3}{q^3}}} = {q^3} - {p^3}$

Q.3 Work out the following divisions.
(i) $(10x - 25) \div 5$
(ii)
$(10x - 25) \div (2x - 5)$
(iii) $10y(6y + 21) \div 5(2y + 7)$
(iv)
$9{x^2}{y^2}(3z - 24) \div 27xy(z - 8)$
(v) $96abc(3a - 12)(5b - 30) \div 144(a - 4)(b - 6)$
Sol. (i) $(10x - 25) \div 5$
= $(10x - 25) \div 5 = {{10x - 25} \over 5}$
$= {{5(2x - 5)} \over 5} = 2x - 5$

(ii)$(10x - 25) \div (2x - 5)$
= $(10x - 25) \div (2x - 5) = {{10x - 25} \over {2x - 5}}$
$= {{5(2x - 5)} \over {(2x - 5)}} = 5$

(iii) $10y(6y + 21) \div 5(2y + 7)$
= $10y(6y + 21) \div 5(2y + 7) = {{10y(6y + 21)} \over {5(2y + 7)}}$
=${{2 \times 5 \times y \times 3(2y + 7)} \over {5(2y + 7)}}$
= $2 \times y \times 3 = 6y$

(iv) $9{x^2}{y^2}(3z - 24) \div 27xy(z - 8)$
= ${{9{x^2}{y^2}(3z - 24)} \over {27xy(z - 8)}}$
= ${9 \over {27}} \times {{x \times y \times x \times y \times 3(z - 8)} \over {x \times y(z - 8)}} = xy$

(v) $96abc(3a - 12)(5b - 30) \div 144(a - 4)(b - 6)$
= $96abc(3a - 12)(5b - 30) \div 144(a - 4)(b - 6) = {{96abc(3a - 12)(5b - 30)} \over {144(a - 4)(b - 6)}}$
= ${{12 \times 4 \times 2 \times a \times b \times c \times 3(a - 4) \times 5(b - 6)} \over {12 \times 4 \times 3(a - 4)(b - 6)}}$
= $10abc$

Q.4 Divide as directed.
(i) $5(2x + 1)(3x + 5) \div (2x + 1)$
(ii)
$26xy(x + 5)(y - 4) \div 13x(y - 4)$
(iii) $52pqr(p + q)(q + r)(r + p) \div 104pq(q + r)(r + p)$
(iv) $20(y + 4)({y^2} + 5y + 3) \div 5(y + 4)$
(v)
$x(x + 1)(x + 2)(x + 3) \div x(x + 1)$
Sol. (i) $5(2x + 1)(3x + 5) \div (2x + 1)$
= $5(2x + 1)(3x + 5) \div (2x + 1) = {{5(2x + 1)(3x + 5)} \over {(2x + 1)}}$
= $5(3x + 1)$

(ii) $26xy(x + 5)(y - 4) \div 13x(y - 4)$
= $26xy(x + 5)(y - 4) \div 13x(y - 4) = {{26xy(x + 5)(y - 4)} \over {13x(y - 4)}}$
= ${{2 \times 13 \times x \times y(x + 5)(y - 4)} \over {13x(y - 4)}}$
= $2y(x + 5)$

(iii) $52pqr(p + q)(q + r)(r + p) \div 104pq(q + r)(r + p)$
= ${{2 \times 2 \times 13 \times p \times q \times r \times (p + q) \times (q + r) \times (r + p)} \over {2 \times 2 \times 2 \times 13 \times p \times q \times (q + r) \times (r + p)}}$
=${1 \over 2}r(p + q)$

(iv) $20(y + 4)({y^2} + 5y + 3) \div 5(y + 4)$
= ${{20(y + 4)({y^2} + 5y + 3)} \over {5(y + 4)}}$
= ${{2 \times 2 \times 5 \times (y + 4) \times ({y^2} + 5y + 3)} \over {5 \times (y + 4)}}$
= $4({y^2} + 5y + 3)$

(v) $x(x + 1)(x + 2)(x + 3) \div x(x + 1)$
= ${{x(x + 1)(x + 2)(x + 3)} \over {x(x + 1)}}$
= $(x + 2)(x + 3)$

Q.5 Factorise the expressions and divide them as directed.
(i) $({y^2} + 7y + 10) \div (y + 5)$
(ii)
$({m^2} - 14m - 32) \div (m + 2)$
(iii) $(5{p^2} - 25p + 20) \div (p - 1)$
(iv)
$4yz({z^2} + 6z - 16) \div 2y(z + 8)$
(v) $5pq({p^2} - {q^2}) \div 2p(p + q)$
(vi)
$12xy(9{x^2} - 16{y^2}) \div 4xy(3x + 4y)$
(vii) $39{y^3}(50{y^2} - 98) \div 26{y^2}(5y + 7)$
Sol. (i) $({y^2} + 7y + 10) \div (y + 5)$
= ${{({y^2} + 7y + 10)} \over {(y + 5)}}$
= ${{{y^2} + 2y + 5y + 2 \times 5} \over {(y + 5)}}$
= ${{(y + 2)(y + 5)} \over {(y + 5)}}$
= $(y + 2)$

(ii) $({m^2} - 14m - 32) \div (m + 2)$
= ${{({m^2} - 14m - 32)} \over {(m + 2)}}$
= ${{{m^2} - 16m + 2m + ( - 16) \times 2} \over {(m + 2)}}$
= ${{(m - 16)(m + 2)} \over {(m + 2)}}$
= $(m + 2)$

(iii) $(5{p^2} - 25p + 20) \div (p - 1)$
= ${{(5{p^2} - 25p + 20)} \over {(p - 1)}}$
= ${{5{p^2} - 20p - 5p + 20} \over {(p - 1)}}$
= ${{5p(p - 4) - 5(p - 4)} \over {(p - 1)}}$
= ${{(5p - 5)(p - 4)} \over {(p - 1)}} = {{5(p - 1)(p - 4)} \over {(p - 1)}}$
= $5(p - 4)$

(iv) $4yz({z^2} + 6z - 16) \div 2y(z + 8)$
= ${{4yz({z^2} + 6z - 16)} \over {2y(z + 8)}}$
= ${{4yz(z + 8z - 2z + 8 \times ( - 2))} \over {2y(z + 8)}}$
= ${{4yz(z - 2)(z + 8)} \over {2y(z + 8)}}$
= $2z(z - 2)$

(v) $5pq({p^2} - {q^2}) \div 2p(p + q)$
= ${{5pq({p^2} - {q^2})} \over {2p(p + q)}}$
= ${{5pq(p - q)(p + q)} \over {2p(p + q)}}$
= ${5 \over 2}q(p - q)$

(vi) $12xy(9{x^2} - 16{y^2}) \div 4xy(3x + 4y)$
= ${{12xy(9{x^2} - 16{y^2})} \over {4xy(3x + 4y)}}$
= ${{12xy({{(3x)}^2} - {{(4y)}^2})} \over {4xy(3x + 4y)}}$
= ${{12xy(3x + 4y)(3x - 4y)} \over {4xy(3x + 4y)}}$
= $3(3x - 4y)$

(vii) $39{y^3}(50{y^2} - 98) \div 26{y^2}(5y + 7)$
= ${{39{y^3}(50{y^2} - 98)} \over {26{y^2}(5y + 7)}}$
= ${{39{y^3} \times 2(25{y^2} - 49)} \over {26{y^2}(5y + 7)}}$
= ${{39{y^3} \times 2[{{(5y)}^2} - {{(7)}^2}]} \over {26{y^2}(5y + 7)}}$
= ${{39{y^3} \times 2 \times (5y - 7)(5y + 7)} \over {26{y^2}(5y + 7)}}$
= $3y(5y - 7)$