• # question_answer 19)                 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)$ $={{x}^{2}}{{y}^{2}}{{z}^{2}}-6xyz\,+8$.

You need to login to perform this action.
You will be redirected in 3 sec