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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)



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