a) \( f(x)=x^{2}+3 x \) \( g(x)=3 x-2 \) 1) \( 3 f(x)-2 g(x) \) i1) \( \frac{12 f(2)}{4 f(1)} \) iii) \( f(4) g(2) \)

To tackle this problem, we first compute \( f(x) \) and \( g(x) \): 1. \( f(x) = x^2 + 3x \) 2. \( g(x) = 3x - 2 \) Now, let's go through each part step by step: 1) For \( 3 f(x) - 2 g(x) \): \[ 3 f(x) = 3(x^2 + 3x) = 3x^2 + 9x \] \[ 2 g(x) = 2(3x - 2) = 6x - 4 \] Now subtract: \[ 3 f(x) - 2 g(x) = (3x^2 + 9x) - (6x - 4) = 3x^2 + 3x + 4 \] i1) Now for \( \frac{12 f(2)}{4 f(1)} \): \[ f(2) = (2^2) + 3(2) = 4 + 6 = 10 \] \[ f(1) = (1^2) + 3(1) = 1 + 3 = 4 \] Then compute: \[ \frac{12 f(2)}{4 f(1)} = \frac{12 \cdot 10}{4 \cdot 4} = \frac{120}{16} = 7.5 \] iii) Lastly, for \( f(4) g(2) \): \[ f(4) = (4^2) + 3(4) = 16 + 12 = 28 \] \[ g(2) = 3(2) - 2 = 6 - 2 = 4 \] So we multiply: \[ f(4) g(2) = 28 \cdot 4 = 112 \] In summary: 1) \( 3 f(x) - 2 g(x) = 3x^2 + 3x + 4 \) i1) \( \frac{12 f(2)}{4 f(1)} = 7.5 \) iii) \( f(4) g(2) = 112 \)