4.(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 (iii) a = -1

(a) Given the expression: \(3x(4x – 5) + 3 = 4x \times 3x – 5 \times 3x + 3 = 12x^2 – 15x + 3\)
(i) So for x = 3, we have

\(12 \times (3)2 – 15 \times 3 + 3 = 12 \times 9 – 45 + 3 = 108 – 42 = 66\)

(ii)For x=\(\frac{1}2\), we have:

\(\Rightarrow 12(\frac{1}2)^2 - 15(\frac{1}2)+3\)

\(\Rightarrow 12\times(\frac{1}4) - \frac{15}2+3\)

\(\Rightarrow 3 - \frac{15}2+3\)

\(\Rightarrow \frac{6-15+6}2=\frac{12-15}2=-\frac{3}2\)

(b) We have \(a(a^2 + a + 1) + 5\)

\(= (a^2 \times a) + (a \times a) + (1 \times a) + 5\)

\(= a^3 + a^2 + a + 5\)

(i) For a = 0, we have:

\(= (0)^3 + (0)^2 + (0) + 5 = 5\)

(ii) For a = 1, we have:

\(= (1)^3 + (1)^2 + (1) + 5 = 1 + 1 + 1 + 5 = 8\)

(iii) For a = -1, we have:

\(= (-1)^3 + (-1)^2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4\)