• # question_answer 13)                 (a) Simplify $3x(4x-5)\,+3$ and find its values for                 (i) $x=3,$                 (ii) $x=\frac{1}{2}$. (b) Simplify: $a({{a}^{2}}+a+1)\,+5$ and find its value for (i) $a=0,$ (ii) a = 1 and ( (iii) a = - 1.

(a) $3x(4x-5)+3$                 $=(3x)\,(4x)-(3x)(5)+3$                 $=(3\times 4)\,\times (x\times x)\,-15x+3$                 (i) When $x=3,\,12{{x}^{2}}-15x+3$                 $=12{{(3)}^{2}}-15(3)+3$ $=108-45+3$ = 66 (ii) When $x=\frac{1}{2},\,12{{x}^{2}}-15x+3$ $=12{{\left( \frac{1}{2} \right)}^{2}}-15\left( \frac{1}{2} \right)+3$ $=3-\frac{15}{2}+3$ $=-\frac{3}{2}$ (b) $a({{a}^{2}}+a+1)+5$                 $=a\times {{a}^{2}}+a\times a+a\times 1+5$ $={{a}^{3}}+{{a}^{2}}+a+5$                 (ii) When a = 1 ${{a}^{3}}+{{a}^{2}}+a+5={{(1)}^{3}}+{{(1)}^{2}}+(1)+5$ $=1+1+1+5$ = 8                 (iii) When a = - 1 ${{a}^{3}}+{{a}^{2}}+a+5$ $={{(-1)}^{3}}+{{(-1)}^{2}}+(-1)+5$ $=-1+1-1+5$ = 8                 (iii) When a = - 1                 ${{a}^{3}}+{{a}^{2}}+a+5$ $={{(-1)}^{3}}+{{(-1)}^{2}}+(-1)+5$  $=-1+1-1+5$ = 4.