question_answer 5)

                Factorize:                 (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) \[25{{a}^{2}}-4{{b}^{2}}+28bc-49{{c}^{2}}\].

Answer:

                (i) \[4{{p}^{2}}-9{{q}^{2}}\] \[4{{p}^{2}}-9{{q}^{2}}={{(2p)}^{2}}-{{(3q)}^{2}}\]                            \[=(2p-3q)\,(2p+3q)\]                                    |Using Identity III (ii) \[63{{a}^{2}}-112{{b}^{2}}\] \[63{{a}^{2}}-112{{b}^{2}}=7\,(9{{a}^{2}}-16{{b}^{2}})\] \[=7\,\{{{(3a)}^{2}}-{{(4b)}^{2}}\}\] \[=7\,(3a-4b)\,(3a+4b)\]                                               |Applying Identity III (iii) \[49{{x}^{2}}-36\] \[49{{x}^{2}}-36\]                            \[={{(7x)}^{2}}-{{(6)}^{2}}\]         \[=(7x-6)\,(7x+6)\]                                          |Using Identity III (iv) \[16{{x}^{5}}-144{{x}^{3}}\] \[16{{x}^{5}}-144{{x}^{3}}\] \[=16{{x}^{3}}\,({{x}^{2}}-9)\] \[=16{{x}^{3}}\,\{{{(x)}^{2}}-{{(3)}^{2}}\}\] \[=16{{x}^{3}}\,(x-3)\,(x+3)\]                                     |Using Identity III (v) \[{{(l+m)}^{2}}-{{(l-m)}^{2}}\] \[{{(l+m)}^{2}}-{{(l-m)}^{2}}\]    \[=\{(l+m)\,-(l-m)\,\}\,\{(l+m)+(l-m)\}\]               |Applying Identity III \[=(2m)\,(2l)\] \[=4\,lm\] (vi) \[9{{x}^{2}}{{y}^{2}}-16\]      \[9{{x}^{2}}{{y}^{2}}-16={{(3xy)}^{2}}-{{(4)}^{2}}\] \[=(3xy-4)\,(3xy+4)\]                                     |Using Identity III (vii) \[({{x}^{2}}-2xy+{{y}^{2}})-{{z}^{2}}\] \[({{x}^{2}}-2xy+{{y}^{2}})-{{z}^{2}}\] \[={{(x-y)}^{2}}-{{z}^{2}}\]                                                           |Using Identity II \[=(x-y-z)\,(x-y+z)\]                                       |Using Identity III (viii) \[25{{a}^{2}}-4{{b}^{2}}+28bc-49{{c}^{2}}\].               \[25{{a}^{2}}-4{{b}^{2}}+28bc-49{{c}^{2}}\] \[=25{{a}^{2}}-(4{{b}^{2}}-28bc+49{{c}^{2}})\] \[=25{{a}^{2}}-\{{{(2b)}^{2}}-2(2b)\,(7c)\,+{{(7c)}^{2}}\}\] \[={{(5a)}^{2}}-{{(2b-7c)}^{2}}\]                                                |Using Identity II \[=\{5a-(2b-7c)\}\,\,\{5a+(2b-7c)\}\] \[=(5a-2b+7c)\,(5a+2b-7c)\].