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Factorisation : Exercise 14.2 (Mathematics NCERT Class 8th)


Q.1 Factorise the following expressions.
(i) {a^2} + 8a + 16
(ii)
{p^2} - 10p + 25
(iii)
25{m^2} + 30m + 9
(iv) 49{y^2} + 84yz + 36{z^2}
(v)
4{x^2} - 8x + 4
(vi)
121{b^2} - 88bc + 16{c^2}
(vii) {(l + m)^2} - 4lm (Hint:Expand {(l + m)^2}first)
(viii) {a^4} + 2{a^2}{b^2} + {b^4}
Sol. (i) {a^2} + 8a + 16
= {(a)^2} + 2 \times a \times 4 + {(4)^2}
= {(a + 4)^2} (Using identity {(a + b)^2} = {a^2} + 2ab + b{}^2)

(ii) {p^2} - 10p + 25
= {(p)^2} - 2 \times p \times 5 + {(5)^2}
= {(p - 5)^2} (Using identity {(a - b)^2} = {a^2} - 2ab + b{}^2)

(iii) 25{m^2} + 30m + 9
= {(5m)^2} + 2 \times 5m \times 3 + {(3)^2}
= {(5m + 3)^2} (Using identity {(a + b)^2} = {a^2} + 2ab + b{}^2)

(iv) 49{y^2} + 84yz + 36{z^2}
= {(7y)^2} + 2 \times 7y \times 6z + {(6z)^2}
= {(7y + 6z)^2} (Using identity {(a + b)^2} = {a^2} + 2ab + b{}^2)

(v) 4{x^2} - 8x + 4
= {(2x)^2} - 2 \times 2x \times 2 + {(2)^2}
= {(2x - 2)^2} (Using identity {(a - b)^2} = {a^2} - 2ab + b{}^2)
= {[2(x - 1)]^2}
= {(x - 1)^2}

(vi) 121{b^2} - 88bc + 16{c^2}
= {(11b)^2} - 2 \times 11b \times 4c + {(4c)^2}
= {(11b - 4c)^2} (Using identity {(a - b)^2} = {a^2} - 2ab + b{}^2)

(vii) {(l + m)^2} - 4lm
= {l^2} + 2 \times l \times m + {m^2} - 4lm
= {l^2} - 2lm + {m^2}
= {(l - m)^2} (Using identity {(a - b)^2} = {a^2} - 2ab + b{}^2)

(viii) {a^4} + 2{a^2}{b^2} + {b^4}
= {({a^2})^2} + 2 \times {a^2} \times {b^2} + {({b^2})^2}
= {({a^2} + {b^2})^2} (Using identity {(a + b)^2} = {a^2} + 2ab + b{}^2)

Q.2 Factorise
(i) 4{p^2} - 9{q^2}
(ii)
63{a^2} - 112{b^2}
(iii)
49{x^2} - 36
(iv) 16{x^5} - 144{x^3}
(v)
{(l + m)^2} - {(l - m)^2}
(vi)
9{x^2}{y^2} - 16
(vii) ({x^2} - 2xy + {y^2}) - {z^2}
(viii)
25{a^2} - 4{b^2} + 28bc - 49{c^2}
Sol. (i) 4{p^2} - 9{q^2}
= {(2p)^2} - {(3q)^2}
= (2p + 3q)(2p - 3q) (Using identity {a^2} - b{}^2 = (a + b)(a - b))

(ii) 63{a^2} - 112{b^2}
= 7(9{a^2} - 16{b^2})
= 7[{(3a)^2} - {(4b)^2}]
= 7(3a + 4b)(3a - 4b) (Using identity {a^2} - b{}^2 = (a + b)(a - b))

(iii) 49{x^2} - 36
= {(7x)^2} - {(6)^2}
= (7x - 6)(7x + 6) (Using identity {a^2} - b{}^2 = (a + b)(a - b))

(iv) 16{x^5} - 144{x^3}
= 16{x^3}[{(x)^2} - 9]
= 16{x^3}[{(x)^2} - {(3)^2}]
= 16{x^3}(x - 3)(x + 3) (Using identity {a^2} - b{}^2 = (a + b)(a - b))

(v) {(l + m)^2} - {(l - m)^2}
=[(l + m) - (l - m)][(l + m) - (l - m)]
(Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (l + m - l + m)(l + m - l + m)
= 2m \times 2l
= 4lm

(vi) 9{x^2}{y^2} - 16
= {(3xy)^2} - {(4)^2}
= (3xy - 4)(3xy + 4) (Using identity {a^2} - b{}^2 = (a + b)(a - b))

(vii) ({x^2} - 2xy + {y^2}) - {z^2}
= {(x - y)^2} - {(z)^2} (Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (x - y - z)(x - y + z) (Using identity {a^2} - b{}^2 = (a + b)(a - b))

(viii) 25{a^2} - 4{b^2} + 28bc - 49{c^2}
= 25{a^2} - (4{b^2} - 28bc + 49{c^2})
= {(5a)^2} - [{(2b)^2} - 2 \times 2b \times 7c + {(7c)^2})]
= {(5a)^2} - {(2b - 7c)^2} (Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (5a + (2b - 7c))(5a - (2b - 7c))(Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (5a + 2b - 7c)(5a - 2b + 7c)

Q.3 Factorise the expressions.
(i) a{x^2} + bx
(ii)
7{p^2} + 21{q^2}
(iii)
2{x^3} + 2x{y^2} + 2x{z^2}
(iv) a{m^2} + b{m^2} + b{n^2} + a{n^2}
(v)
(lm + l) + m + 1
(vi) y(y + z) + 9(y + z)
(vii)
5{y^2} - 20y - 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix)
6xy - 4y + 6 - 9x
Sol. (i) a{x^2} + bx
a{x^2} + bx = a \times x \times x + b \times x
=x(ax + b)

(ii) 7{p^2} + 21{q^2}
7{p^2} + 21{q^2} = 7 \times p \times p + 21 \times q \times q
=7({p^2} + 3{q^2})

(iii) 2{x^3} + 2x{y^2} + 2x{z^2}
= 2x({x^2} + {y^2} + {z^2})

(iv) a{m^2} + b{m^2} + b{n^2} + a{n^2}
a{m^2} + b{m^2} + b{n^2} + a{n^2}
= {m^2}(a + b) + {n^2}(a + b)
= ({m^2} + {n^2})(a + b)

(v) (lm + l) + m + 1
= lm + m + l + 1
= m(l + 1) + 1(l + 1)
= (l + 1)(m + 1)

(vi) y(y + z) + 9(y + z)
= (y + z)(y + 9)

(vii) 5{y^2} - 20y - 8z + 2yz
= 5y(y - 4) + 2z(y - 4)
= (y - 4)(5y + 2z)

(viii) 10ab + 4a + 5b + 2
= 5b(2a + 1) + 2(2a + 1)
= (2a + 1)(5b + 2)

(ix) 6xy - 4y + 6 - 9x
= 3x(2y - 3) - 2(2y - 3)
= (2y - 3)(3x - 2)

Q.4 Factorise.
(i) {a^4} - {b^4}
(ii)
{p^4} - 81
(iii)
{x^4} - {(y + z)^4}
(iv) {x^4} - {(x - z)^4}
(v)
{a^4} - 2{a^2}{b^2} + {b^4}
Sol. (i) {a^4} - {b^4}
= {({a^2})^2} - {({b^2})^2}
= ({a^2} - {b^2})({a^2} + {b^2}) (Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (a - b)(a + b)({a^2} + {b^2}) (Using identity {a^2} - b{}^2 = (a + b)(a - b))

(ii) {p^4} - 81
= {({p^2})^2} - {(9)^2}
= ({p^2} - 9)({p^2} + 9) (Using identity {a^2} - b{}^2 = (a + b)(a - b))\
= ({p^2} - {3^2})({p^2} + 9)(Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (p - 3)(p + 3)({a^2} + {b^2})

(iii) {x^4} - {(y + z)^4}
= {({x^2})^2} - {[{(y + z)^2}]^2}
= ({x^2} - {(y + z)^2})({x^2} + {(y + z)^2}) (Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (x - (y + z))(x + (y + z))({x^2} + {(y + z)^2})(Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (x - y + z)(x + y + z)({x^2} + {(y + z)^2})

(iv) {x^4} - {(x - z)^4}
= {({x^2})^2} - {[{(x - z)^2}]^2}
= ({x^2} - {(x - z)^2})({x^2} + {(x - z)^2}) (Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (x - (x - z))(x + (x + z))({x^2} + {(x - z)^2})(Using identity {a^2} - b{}^2 = (a + b)(a - b))
= (x - x + z)(x + x + z)({x^2} + {x^2} - 2xz + {z^2})
= x(2x + z)(2{x^2} - 2xz + {z^2})

(v) {a^4} - 2{a^2}{b^2} + {b^4}
= {({a^2})^2} - 2{a^2}{b^2} + {({b^2})^2}
= {({a^2} - {b^2})^2} (Using identity {a^2} - b{}^2 = (a + b)(a - b))
= {((a - b)(a + b))^2}(Using identity {a^2} - b{}^2 = (a + b)(a - b))
= {(a - b)^2}{(a + b)^2}

Q.5 Factorise the following expressions.
(i) {p^2} + 6p + 8
(ii)
{q^2} - 10q + 21
(iii)
{p^2} + 6p - 16
Sol. (i) {p^2} + 6p + 8
= {p^2} + (4 + 2)p + 4 \times 2
= {p^2} + 4p + 2p + 4 \times 2
= p(p + 4) + 2(p + 4)
= (p + 4)(p + 2)

(ii) {q^2} - 10q + 21
= {q^2} - (7 + 3)q + 7 \times 3
= {q^2} - 7q - 3q + 7 \times 3
= q(q - 7) - 3(q - 7)
= (q - 7)(q - 3)



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