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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:

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