question_answer 1)

                Use the identity \[(x+a)\,(x+b)\,={{x}^{2}}\]\[+(a+b)\,x+ab\]   to find the following products:                 (i) \[(x+3)\,(x+7)\]                 (ii) \[(4x+5)\,(4x+1)\]                 (iii) \[(4x-5)\,(4x-1)\]                 (iv) \[(4x+5)\,(4x-1)\]                 (v) \[(2x+5y)\,(2x+3y)\]                 (vi) \[(2{{a}^{2}}+9)\,(2{{a}^{2}}+5)\]                 (vii) \[(xyz-4)\,(xyz\,-2)\]

Answer:

                (i) \[(x+3)\,(x+7)\]           \[(x+3)\,(x+7)\] \[={{x}^{2}}+(3+7)x+(3)\,(7)\] \[={{x}^{2}}+10x+21\] (ii) \[(4x+5)\,(4x+1)\] \[(4x+5)\,(4x+1)\,={{(4x)}^{2}}+(5+1)(4x)\]\[+(5)\,(1)\] \[=16{{x}^{2}}+24x+5\] (iii) \[(4x-5)\,(4x-1)\] \[(4x-5)\,(4x-1)\] \[=\{4x+(-5)\,\}\,\{4x+(-1)\}\] \[={{(4x)}^{2}}+\{(-5)+(-1)\}\,(4x)\,+(-5)\,(-1)\] \[=16{{x}^{2}}-2x+5\] (iv) \[(4x+5)\,(4x-1)\] \[(4x+5)\,(4x-1)\]\[=(4x+5)\,\{4x+(-1)\}\] \[={{(4x)}^{2}}+\{5+(-1)\}\,(4x)+(5)(-1)\] \[=16{{x}^{2}}+16x-5\] (v) \[(2x+5y)\,(2x+3y)\] \[(2x+5y)\,(2x+3y)\] \[={{(2x)}^{2}}+(5y+3y)\,(2x)\,+(5y)\,(3y)\] \[=4{{x}^{2}}+(8y)\,(2x)\,+15{{y}^{2}}\] \[=4{{x}^{2}}\,+16xy+15{{y}^{2}}\] (vi) \[=(2{{a}^{2}}+9)\,(2{{a}^{2}}+5)\] \[=(2{{a}^{2}}+9)\,(2{{a}^{2}}+5)\] \[={{(2{{a}^{2}})}^{2}}+(5+9)\,{{(2a)}^{2}}+(5)\,(9)\] \[=4{{a}^{4}}+28{{a}^{2}}+45\] (vii) \[(xyz-4)\,(xyz-2)\] \[(xyz-4)\,(xyz-2)\] \[=\{xyz+(-4)\}\,\{xyz\,+(-2)\}\] \[={{(xyz)}^{2}}\,+\{(-4)\,+(-2)\}\,(xyz)\]\[+(-4)\,(-2)\] \[={{x}^{2}}{{y}^{2}}{{z}^{2}}-6xyz\,+8\].