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

1 Comment

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