Answer:
(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.
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