# Polynomials : Exercise 2.5 (Mathematics NCERT Class 9th)

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Q.1     Use suitable identities to find the following products :
(i) (x + 4) (x + 10)                   (ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)                (iv) $\left( {{y^2} + {3 \over 2}} \right)\left( {{y^2} - {3 \over 2}} \right)$

(v) (3 – 2x) (3 + 2x)
Sol.

(i) (x + 4) (x + 10) = ${x^2} + \left( {4 + 10} \right)x + 4 \times 10$
$= {x^2} + 14x + 40$
(ii) (x + 8) (x –10)  $= {x^2} + \left( {8 - 10} \right)x + 8 \times (- 10)$

$= {x^2} - 2x - 80$
(iii) (3x + 4) (3x – 5) = $3x\left( {3x - 5} \right) + 4\left( {3x - 5} \right)$

$= 3x \times 3x - 3x \times 5 + 4 \times 3x-4 \times 5$
$= 9{x^2} - 15x + 12x - 20$
$= 9{x^2} - 3x - 20$
(iv) $\left( {{y^2} + {3 \over 2}} \right)\left( {{y^2} - {3 \over 2}} \right)$
$= {\left( {{y^2}} \right)^2} - {\left( {{3 \over 2}} \right)^2}$

$= {y^4} - {9 \over 4}$
(v) (3 – 2x) (3 + 2x) = ${\left( 3 \right)^2} - {\left( {2x} \right)^2}$

$= 9 - 4{x^2}$

Q.2      Evaluate the following products without multiplying directly :

(i) 103 × 107      (ii) 95 × 96       (iii) 104 × 96
Sol.

(i) 103 × 107 = $\left( {100 + 3} \right)\left( {100 + 7} \right)$

= ${\left( {100} \right)^2} + \left( {3 + 7} \right)\left( {100} \right) + 3 \times 7$
= $100 \times 100 + \left( {10} \right)\left( {100} \right) + 21$
= $10000 + 1000 + 21 = 11021$
(ii) 95 × 96 = $\left( {100 - 5} \right)\left( {100 - 4} \right)$
= ${\left( {100} \right)^2} + \left( { - 5 - 4} \right)\left( {100} \right) + \left( { - 5} \right)\left( { - 4} \right)$
= $100 \times 100 + \left( { - 9} \right)\left( {100} \right) + 20$
= $10000 - 900 + 20 = 9120$
(iii) 104 × 96 = $\left( {100 + 4} \right)\left( {100 - 4} \right)$
= ${\left( {100} \right)^2} - {\left( 4 \right)^2}$
= $10000 - 16 = 9984$

Q.3      Factorise the following using appropriate identifies :

(i) $9{x^2} + 6 xy + {y^2}$
(ii) $4{y^2} - 4y + 1$

(iii) ${x^2} - {{{y^2}} \over {100}}$
Sol.

(a) $9{x^2} + 6 xy + {y^2}$ = ${\left( {3x} \right)^2} + 2\left( {3x} \right)\left( y \right) + {\left( y \right)^2}$ $\left[ {{{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}} \right]$
= ${\left( {3x + y} \right)^2} = \left( {3x + y} \right)\left( {3x + y} \right)$

(b) $4{y^2} - 4y + 1$ = ${\left( {2y} \right)^2} - 2\left( {2y} \right)\left( 1 \right) + {\left( 1 \right)^2}$ $\left[ {{{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right]$
= ${\left( {2y - 1} \right)^2} = \left( {2y - 1} \right)\left( {2y - 1} \right)$
(c) ${x^2} - {{{y^2}} \over {100}}$ = ${\left( x \right)^2} - {\left( {{y \over {10}}} \right)^2}$ $\left[ {\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}} \right]$
$= \left( {x - {y \over {10}}} \right)\left( {x + {y \over {10}}} \right)$

Q.4      Expand each of the following using suitable identifies :

(i) ${\left( {x + 2y + 4z} \right)^2}$

(ii) ${\left( {2x - y + z} \right)^2}$

(iii) ${\left( { - 2x + 3y + 2z} \right)^2}$

(iv) ${\left( {3a - 7b - c} \right)^2}$

(v) ${\left( { - 2x + 5y - 3z} \right)^2}$

(vi) ${\left[ {{1 \over 4}a - {1 \over 2}b + 1} \right]^2}$
Sol.

(i) ${\left( {x + 2y + 4z} \right)^2}$
$= {x^2} + {\left( {2y} \right)^2} + {\left( {4z} \right)^2} + 2\left( x \right)\left( {2y} \right)$$+ 2\left( {2y} \right)\left( {4z} \right) + 2\left( {4z} \right)\left( x \right)$
$= {x^2} + 4{y^2} + 16{z^2} + 4xy + 16yz + 8zx$

(ii) ${\left( {2x - y + z} \right)^2}$
$= {\left[ {2x + \left( { - y} \right) + z} \right]^2}$
$= {\left( {2x} \right)^2} + {\left( { - y} \right)^2} + {z^2} + 2\left( {2x} \right)\left( { - y} \right)$$+ 2\left( { - y} \right)\left( z \right) + 2\left( z \right)\left( {2x} \right)$
$= 4{x^2} + {y^2} + {z^2} - 4xy - 2yz + 4zx$

(iii) ${\left( { - 2x + 3y + 2z} \right)^2}$
$= {\left[ {\left( { - 2x} \right) + 3y + 2z} \right]^2}$
$= {\left( { - 2x} \right)^2} + {\left( {3y} \right)^2} + {\left( {2z} \right)^2}$ $+ 2\left( { - 2x} \right)\left( {3y} \right) + 2\left( {3y} \right)\left( {2z} \right)$$+ 2\left( {2z} \right)\left( { - 2x} \right)$
$= 4{x^2} + 9{y^2} + 4{z^2} - 12xy + 12yz - 8zx$

(iv) ${\left( {3a - 7b - c} \right)^2}$
$= {\left[ {3a + \left( { - 7b} \right) + \left( { - c} \right)} \right]^2}$
$= {\left( {3a} \right)^2} + {\left( { - 7b} \right)^2} + {\left( { - c} \right)^2}$$+ 2\left( {3a} \right)\left( { - 7b} \right) + 2\left( { - 7b} \right)\left( { - c} \right)$$+ 2\left( { - c} \right)\left( {3a} \right)$
$= 9{a^2} + 49{b^2} + {c^2} - 42\,ab + 14\,bc - 6ca$

(v) ${\left( { - 2x + 5y - 3z} \right)^2}$
$= {\left[ {{{\left( { - 2x} \right)}^2} + 5y + \left( { - 3z} \right)} \right]^2}$
$= {\left( { - 2x} \right)^2} + {\left( {5y} \right)^2} + {\left( { - 3z} \right)^2}$$+ 2\left( { - 2x} \right)\left( {5y} \right) + 2\left( {5y} \right)\left( { - 3z} \right)$$+ 2\left( { - 3z} \right)\left( { - 2x} \right)$
$= 4{x^2} + 25{y^2} + 9{z^2} - 20xy - 30yz + 12zx$

(vi) ${\left[ {{1 \over 4}a - {1 \over 2}b + 1} \right]^2}$
$= {\left[ {{1 \over 4}a + \left( {{-1 \over 2}b} \right) + 1} \right]^2}$
$= {\left( {{1 \over 4}a} \right)^2} + {\left( { - {1 \over 2}b} \right)^2}$$+ {\left( 1 \right)^2} + 2\left( {{1 \over 4}a} \right)\left( { - {1 \over 2}b} \right)$$+ 2\left( { - {1 \over 2}b} \right)\left( 1 \right) + 2\left( 1 \right)\left( {{1 \over 4}a} \right)$
$= {1 \over {16}}{a^2} + {1 \over 4}{b^2} + 1 - {1 \over 4}ab - b + {1 \over 2}a$

Q.5      Factorise :
(i) $4{x^2} + 9{y^2} + 16{z^2} + 12xy - 24yz - 16xz$
(ii) $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy - 4\sqrt 2 \,yz - 8xz$
Sol.

(i) $4{x^2} + 9{y^2} + 16\,{z^2} + 12\,xy - 24\,yz - 16\,xz$
$= {\left( {2x} \right)^2} + {\left( {3y} \right)^2} + {\left( { - 4{\mkern 1mu} z} \right)^2}$$+ 2\left( {2x} \right)\left( {3y} \right) + 2\left( {3y} \right)\left( { - 4z} \right)$$+ 2\left( {2x} \right)\left( { - 4z} \right)$
= ${\left[ {2x + 3y + \left( { - 4z} \right)} \right]^2} = {\left( {2x + 3y - 4z} \right)^2}$

(ii) $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
$= {\left( {\sqrt 2 x} \right)^2} + {\left( { - y} \right)^2} + {\left( { - 2\sqrt 2 {\mkern 1mu} z} \right)^2}$$+ 2\left( {\sqrt {2x} } \right)\left( { - y} \right) + 2\left( { - y} \right)\left( { - 2\sqrt 2 {\mkern 1mu} z} \right)$$+ 2\left( {\sqrt {2x} } \right)\left( { - 2\sqrt {2z} } \right)$
$= {\left[ {\sqrt 2 x + \left( { - y} \right) + \left( { - 2\sqrt 2 z} \right)} \right]^2}$
$= {\left( {\sqrt 2 x - y - 2\sqrt 2 z } \right)^2}$

Q.6      Write the following cubes in expanded form :
(i) ${\left( {2x + 1} \right)^3}$
(ii) ${\left( {2a - 3b} \right)^3}$
(iii) ${\left[ {{3 \over 2}x + 1} \right]^3}$
(iv) ${\left[ {x - {2 \over 3}y} \right]^3}$
Sol.

(i) ${\left( {2x + 1} \right)^3}$ = ${\left( {2x} \right)^3} + 3{\left( {2x} \right)^2}\left( 1 \right) + 3\left( {2x} \right){\left( 1 \right)^2} + {\left( 1 \right)^3}$ $= 8{x^3} + 12{x^2} + 6x + 1$

(ii) ${\left( {2a - 3b} \right)^3}$ = ${\left( {2a} \right)^3} - 3{\left( {2a} \right)^2}\left( {3b} \right) + 3\left( {2a} \right){\left( {3b} \right)^2} - {\left( {3b} \right)^3}$ = $8{a^3} - 36\,{a^2}b + 54\,a{b^2} - 27\,{b^3}$

(iii)${\left[ {{3 \over 2}x + 1} \right]^3}$$= {\left( {{3 \over 2}x} \right)^3} + 3{\left( {{3 \over 2}x} \right)^2}\left( 1 \right) + 3\left( {{3 \over 2}x} \right){\left( 1 \right)^2} + {1^3}$ $= {{27} \over 8}{x^3} + {{27} \over 4}{x^2} + {9 \over 2}x + 1$

(iv) ${\left[ {x - {2 \over 3}y} \right]^3}$$= {x^3} - 3{\left( x \right)^2}\left( {{2 \over 3}y} \right) + 3\left( x \right){\left( {{2 \over 3}y} \right)^2} - {\left( {{2 \over 3}y} \right)^3}$ $= {x^3} - 2{x^2}y + {4 \over 3}x{y^2} - {8 \over {27}}{y^3}$

Q.7      Evaluate the following using suitable identities
(i) ${\left( {99} \right)^3}$
(ii) ${\left( {102} \right)^3}$
(iii) ${\left( {998} \right)^3}$
Sol.

(i) ${\left( {99} \right)^3}$$= {\left( {100 - 1} \right)^3}$
$= {\left( {100} \right)^3} - {1^3} - 3\left( {100} \right)\left( 1 \right)\left( {100 - 1} \right)$

$= 1000000 - 1 - 29700 = 970299$

(ii) ${\left( {102} \right)^3}$$= {\left( {100 + 2} \right)^3}$
${\left( {100} \right)^3} + {\left( 2 \right)^3} + 3\left( {100} \right)\left( 2 \right)\left( {100 + 2} \right)$
$= 1000000 + 8 + 61200 = 1061208$

(iii) ${\left( {998} \right)^3}$$= {\left( {1000 - 2} \right)^3}$
$= {\left( {1000} \right)^3} - {\left( 2 \right)^3} - 3\left( {1000} \right)\left( 2 \right)\left( {1000 - 2} \right)$

$= 1000000000 - 8 - 5988000$$= 994011992$

Q.8      Factorise each of the following :
(i) $8{a^3} + {b^3} + 12{a^2}b + 6a{b^2}$
(ii) $8{a^3} - {b^3} - 12{a^2}b + 6a{b^2}$
(iii) $27 - 125{a^3} - 135a + 225{a^2}$
(iv)$64\,{a^3} - 27\,{b^3} - 144\,{a^2}b + 108\,a{b^2}$
(v) $27{p^3} - {1 \over {216}} - {9 \over 2}{p^2} + {1 \over 4}p$
Sol.

(i) $8{a^3} + {b^3} + 12{a^2}b + 6a{b^2}$
$= {\left( {2a} \right)^3} + {\left( b \right)^3} + 3\left( {2a} \right)\left( b \right)\left( {2a + b} \right)$
$= {\left( {2a + b} \right)^3}$
$= \left( {2a + b} \right)\left( {2a + b} \right)\left( {2a + b} \right)$

(ii) $8{a^3} - {b^3} - 12{a^2}b + 6a{b^2}$
$= {\left( {2a} \right)^3} - {b^3} - 3\left( {2a} \right)\left( b \right)\left( {2a - b} \right)$
$= {\left( {2a - b} \right)^3}$
$= \left( {2a - b} \right)\left( {2a - b} \right)\left( {2a - b} \right)$

(iii) $27 - 125{a^3} - 135a + 225{a^2}$
$= {\left( 3 \right)^3} - {\left( {5a} \right)^3} - 3\left( 3 \right)\left( {5a} \right)\left( {3 - 5a} \right)$
$= {\left( {3 - 5a} \right)^3}$
$= \left( {3 - 5a} \right)\left( {3 - 5a} \right)\left( {3 - 5a} \right)$

(iv) $64\,{a^3} - 27\,{b^3} - 144\,{a^2}b + 108\,a{b^2}$
$= {\left( {4a} \right)^3} - {\left( {3b} \right)^3} - 3\left( {4a} \right)\left( {3b} \right)\left( {4a - 3b} \right)$
$= {\left( {4a - 3b} \right)^3}$
$= \left( {4a - 3b} \right)\left( {4a - 3b} \right)\left( {4a - 3b} \right)$

(v) $27{p^3} - {1 \over {216}} - {9 \over 2}{p^2} + {1 \over 4}p$
$= {\left( {3p} \right)^3} - {\left( {{1 \over 6}} \right)^3} - 3\left( {3p} \right)\left( {{1 \over 6}} \right)\left( {3p - {1 \over 6}} \right)$
$= {\left( {3p - {1 \over 6}} \right)^3} = \left( {3p - {1 \over 6}} \right)\left( {3p - {1 \over 6}} \right)\left( {3p - {1 \over 6}} \right)$

Q.9      Verify :
(i) ${x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)$

(ii) ${x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)$
Sol.

(i) R.H.S $= \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)$
$= x\left( {{x^2} - xy + {y^2}} \right) + y\left( {{x^2} - xy + {y^2}} \right)$
$= {x^3} - {x^2}y + x{y^2} + {x^2}y - x{y^2} + {y^3}$
$= {x^3} + {y^3} = L.H.S.$
Thus, verified.
(ii) R.H.S. $= \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)$

$= x\left( {{x^2} + xy + {y^2}} \right) - y\left( {{x^2} + xy + {y^2}} \right)$
$= {x^3} + {x^2}y + x{y^2} - {x^2}y - x{y^2} - {y^3}$
${x^3} - {y^3} = L.H.S.$
Thus, verified.

Q.10     Factorise each of the following :
(i) $27\,{y^3} + 125\,{z^3}$
(ii) $64\,{m^3} - 343\,{n^3}$
Sol.

(i) $27\,{y^3} + 125\,{z^3}$
$= {\left( {3y} \right)^3} + {\left( {5z} \right)^3}$
$= \left( {3y + 5z} \right)\left[ {{{\left( {3y} \right)}^2} - \left( {3y} \right)5z + {{\left( {5z} \right)}^2}} \right]$
$= \left( {3y + 5z} \right)\left( {9{y^2} - 15\,yz + 25\,{z^2}} \right)$

(ii) $64\,{m^3} - 343\,{n^3}$ = ${\left( {4m} \right)^3} - {\left( {7n} \right)^3}$
$= \left( {4m - 7n} \right)\left[ {{{\left( {4m} \right)}^2} + \left( {4m} \right)\left( {7n} \right) + {{\left( {7n} \right)}^2}} \right]$
$= \left( {4m - 7n} \right)\left( {16{m^2} + 28mn + 49{n^2}} \right)$

Q.11     Factorise : $27{x^3} + {y^3} + {z^3} - 9xyz$
Sol.

$27{x^3} + {y^3} + {z^3} - 9xyz$
$= {\left( {3x} \right)^3} + {y^3} + {z^3} - 3\left( {3x} \right)\left( y \right)\left( z \right)$
$= \left( {3x + y + z} \right)\left[ {{{\left( {3x} \right)}^2} + {y^2} + {z^2} - \left( {3x} \right)y - yz - z\left( {3x} \right)} \right]$
$= \left( {3x + y + z} \right)\left( {9{x^2} + {y^2} + {z^2} - 3xy - yz - 3zx} \right)$

Q.12    Verify that ${x^3} + {y^3} + {z^3} - 3xyz = {1 \over 2}\left( {x + y + z} \right)\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$
Sol.

R.H.S $= {1 \over 2}\left( {x + y + z} \right)\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$
$= {1 \over 2}\left( {x + y + z} \right)\left( {{x^2} - 2xy + {y^2} + {y^2} - 2yz + {z^2} + {z^2} - 2zx + {x^2}} \right)$
$\; = {1 \over 2}(x + y + z)(2x^2+ 2y^2+ 2z^2- 2xy- 2yz- 2zx)$
$= {2 \over 2}(x + y + z)(x^2+ y^2+ z^2- xy- yz- zx)$
$= \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - yz - zx - xy - xy - yz - zx} \right)$

$= {x^3} + {y^3} + {z^2} - 3xyz$
= L.H.S.
Hence verified.

Q.13     If x + y + z = 0 , show that ${x^3} + {y^3} + {z^3} = 3xyz$.
Sol.

We have, x + y + z = 0
$\Rightarrow$ x + y = – z
Cubing both sides, we have
${\left( {x + y} \right)^3} = {\left( { - z} \right)^3}$
$\Rightarrow$ ${x^3} + {y^3} + 3xy\left( {x + y} \right) = - {z^3}$
$\Rightarrow$ ${x^3} + {y^3} - 3xyz = - {z^3}$ [since x + y = – z]
$\Rightarrow$ ${x^3} + {y^3} + {z^3} = 3xyz$, which stands proved.

Q.14     Without actually calculating the cubes, find the value of each of the following :
(i) ${\left( { - 12} \right)^3} + {\left( 7 \right)^3} + {\left( 5 \right)^3}$
(ii) ${\left( {28} \right)^3} + {\left( { - 15} \right)^3} + {\left( { - 13} \right)^3}$
Sol.

(i) Let x = – 12, y = 7 and z = 5
Here $x + y + z = - 12 + 7 + 5 = 0$
$\Rightarrow$ ${x^3} + {y^3} + {z^3} = 3xyz$
$\Rightarrow$ ${\left( { - 12} \right)^3} + {\left( 7 \right)^3} + {\left( 5 \right)^3} = 3 \times \left( { - 12} \right) \times 7 \times 5 = - 1260$

(ii) Let x = 28, y = – 15 and z = – 13
Here, $x + y + z = 28 - 15 - 13 = 0$
$\Rightarrow$ ${x^3} + {y^3} + {z^3} = 3xyz$
$\Rightarrow$ ${\left( {28} \right)^3} + {\left( { - 15} \right)^3} + {\left( { - 13} \right)^3}$
$= 3\left( {28} \right)\left( { - 15} \right)\left( { - 13} \right) = 16380$

Q.15     Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :
(i) $Area\,\,:\,\,25{a^2} - 35a + 12$   (ii) $Area\,\,:\,\,35{y^2} + 13y - 12$
Sol.

Possible length and breadth of the rectangle are the factors of its given area.
Area = length × breadth
(i) $Area\,\,:\,\,25\,{a^2} + 35\,a + 12 = 25\,{a^2} - 15a - 20a + 12$
$= 5a\left( {5a - 3} \right) - 4\left( {5a - 3} \right) = \left( {5a - 3} \right)\left( {5a - 4} \right)$
Therefore Possible length and breadth are (5a – 3) and (5a – 4) units.
(ii) $Area = 35\,{y^2} + 13\,y - 12$

$= 35{y^2} + 28y - 15y - 12$
$= 7y\left( {5y + 4} \right) - 3\left( {5y + 4} \right)$
$= \left( {5y + 4} \right)\left( {7y - 3} \right)$
Therefore possible length and breadth are (5y + 4) and (7y – 3) units.

Q.16     What are the possible expressions for the dimensions of the cuboids whose volumes are given below :

(i) Volume : $3{x^2} - 12x$
(ii) Volume : $12k{y^2} + 8ky - 20k$
Sol.

Possible expressions for the dimensions of the cuboids are the factors of their volumes.
(i) Volume $= 3{x^2} - 12x = 3x\left( {x - 4} \right)$
Therefore Possible dimensions of cuboid are 3, x and (x – 4) units

(ii) Volume $= 12k{y^2} + 8ky - 20\,k$
$= 4k\left( {3{y^2} + 2y - 5} \right)$
$= 4k\left[ {3y\left( {y - 1} \right) + 5\left( {y - 1} \right)} \right]$
$= 4k\left( {y - 1} \right)\left( {3y + 5} \right)$
Therefore Possible dimensions of cuboid are
4k, (y – 1) and (3y + 5) units.

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