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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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