question_answer 1)

                Factorize the expressions and divide them as directed.                 (i) \[({{y}^{2}}+7y+10)\div (y+5)\]                                            (ii) \[({{m}^{2}}-14m-32)\div (m+2)\]                 (iii) \[(5{{p}^{2}}-25p+20)\div (p-1)\]                                       (iv) \[4yz({{z}^{2}}+6z-16)\div \,2y(z+8)\]                 (v) \[5pq({{p}^{2}}-{{q}^{2}})\,\div 2p\,(p+q)\]                                  (vi) \[12xy\,(9{{x}^{2}}-16{{y}^{2}})\div 4xy\,(3x+4y)\]                 (vii) \[39{{y}^{3}}(50{{y}^{2}}-98)\,\div 26{{y}^{2}}(5y+7)\].

Answer:

                (i) \[({{y}^{2}}+7y+10)\div (y+5)\]                 \[({{y}^{2}}+7y+10)\div (y+5)\] \[=\frac{{{y}^{2}}+7y+10}{y+5}\] \[=\frac{{{y}^{2}}+2y+5y+10}{y+5}\]                       |Using Identity IV \[=\frac{y(y+2)+5(y+2)}{y+\,5}\] \[=\frac{(y+2)(y+5)}{y+5}\] \[=y+2\]                 (ii) \[({{m}^{2}}-14m-32)\div (m+2)\]                 \[({{m}^{2}}-14m-32)\div (m+2)\] \[=\frac{{{m}^{2}}-14m-32}{m+2}\] \[=\frac{{{m}^{2}}-16m+2m-32}{m+2}\]                                |Using Identity IV \[=\frac{m(m-16)+2(m-16)}{m+2}\] \[=\frac{(m-16)\,(m+2)}{m+2}\] \[=m-16\]                 (iii) \[(5{{p}^{2}}-25p+20)\div (p-1)\]                 \[(5{{p}^{2}}-25p+20)\div (p-1)\]                 \[=\frac{5({{p}^{2}}-5p+4)}{p-1}\] \[=\frac{5({{p}^{2}}-p-4\,p+4)}{p-1}\]                    |Applying Identity IV \[=\frac{5\{p(p-1)-4(p-1)\}}{p-1}\] \[=\frac{5(p-1)\,(p-4)}{p-1}\] \[=5\,(p-4)\]                 (iv) \[4yz({{z}^{2}}+6z-16)\div \,2y(z+8)\] \[4yz({{z}^{2}}+6z-16)\div \,2y(z+8)\] \[=\frac{4yz({{z}^{2}}+6z-16)}{2y\,(z+8)}\] \[=\frac{2z({{z}^{2}}+6z-16)}{z+8}\] \[=\frac{2z({{z}^{2}}+8z-2z-16)}{z+8}\]                   |Using Identity IV \[=\frac{2z[z(z+8)\,-2(z+8)]}{z+8}\] \[=\frac{2z(z+8)\,(z-2)}{z+8}\] \[=2z\,(z-2)\]                 (v) \[5pq({{p}^{2}}-{{q}^{2}})\,\div 2p\,(p+q)\]                 \[5pq({{p}^{2}}-{{q}^{2}})\,\div 2p\,(p+q)\] \[=\frac{5pq({{p}^{2}}-{{q}^{2}})}{2p\,(p+q)}\] \[=\frac{5\,pq\,(p+q)\,(p-q)}{2p(p+q)}\]                              |Using Identity III \[=\frac{5}{2}q(p-q)\]                 (vi) \[12xy\,(9{{x}^{2}}-16{{y}^{2}})\div 4xy\,(3x+4y)\]                 \[12xy\,(9{{x}^{2}}-16{{y}^{2}})\div 4xy\,(3x+4y)\]                 \[=\frac{12xy(9{{x}^{2}}-16{{y}^{2}})}{4xy(3x+4y)}\]                 \[\frac{3(9{{x}^{2}}-16{{y}^{2}})}{3x+4y}\]                 \[=\frac{3\{{{(3x)}^{2}}-(4{{y}^{2}})\}}{3x+4y}\]                 \[=\frac{3(3x+4y)\,(3x-4y)}{3x+4y}\]                 \[=3(3x-4y)\]                 (vii) \[39{{y}^{3}}(50{{y}^{2}}-98)\,\div 26{{y}^{2}}(5y+7)\].                 \[39{{y}^{3}}(50{{y}^{2}}-98)\,\div 26{{y}^{2}}(5y+7)\] \[=\frac{39{{y}^{3}}\,(50{{y}^{2}}-98)}{26{{y}^{2}}(5y+7)}\] \[=\frac{39{{y}^{3}}\times 2\times (25{{y}^{2}}-49)}{26{{y}^{2}}(5y+7)}\] \[=\frac{39{{y}^{3}}\times 2\times \{{{(5y)}^{2}}-{{(7)}^{2}}\}}{26{{y}^{2}}\,(5y+7)}\] \[=\frac{39{{y}^{3}}\times 2\times (5y+7)\,(5y-7)}{26{{y}^{2}}\,(5y+7)}\]                              |Using Identity III \[=3y(5y-7)\].