The problem has three parts: (i) Simplify the expression $(4\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})$. (ii) Given the function $f(x) = 3x^2 - 4$, find $f'(x)$ from first principles. (iii) Solve the polynomial equation $3x^3 + 8x^2 - 15x + 4 = 0$.

1. Problem Description

The problem has three parts:

(i) Simplify the expression $(4\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})$.

(ii) Given the function $f(x) = 3x^2 - 4$, find $f'(x)$ from first principles.

(iii) Solve the polynomial equation $3x^3 + 8x^2 - 15x + 4 = 0$.

2. Solution Steps

(i) Simplify $(4\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})$:

Expand the expression:

$(4\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3}) = (4\sqrt{2})(3\sqrt{2}) + (4\sqrt{2})(2\sqrt{3}) + (\sqrt{3})(3\sqrt{2}) + (\sqrt{3})(2\sqrt{3})$

$= 12(2) + 8\sqrt{6} + 3\sqrt{6} + 2(3)$

$= 24 + 11\sqrt{6} + 6$

$= 30 + 11\sqrt{6}$

(ii) Find $f'(x)$ from first principles:

$f(x) = 3x^2 - 4$

The definition of the derivative from first principles is:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f(x+h) = 3(x+h)^2 - 4 = 3(x^2 + 2xh + h^2) - 4 = 3x^2 + 6xh + 3h^2 - 4$

$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 4) - (3x^2 - 4) = 6xh + 3h^2$

$\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2}{h} = 6x + 3h$

$f'(x) = \lim_{h \to 0} (6x + 3h) = 6x + 3(0) = 6x$

$f'(x) = 6x$

(iii) Solve the polynomial equation $3x^3 + 8x^2 - 15x + 4 = 0$:

We look for rational roots using the Rational Root Theorem. Possible rational roots are $\pm 1, \pm 2, \pm 4, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}$.

Try $x = 1/3$:

$3(\frac{1}{3})^3 + 8(\frac{1}{3})^2 - 15(\frac{1}{3}) + 4 = 3(\frac{1}{27}) + 8(\frac{1}{9}) - 5 + 4 = \frac{1}{9} + \frac{8}{9} - 1 = \frac{9}{9} - 1 = 1 - 1 = 0$

So $x = \frac{1}{3}$ is a root. This means $(3x-1)$ is a factor.

Divide $3x^3 + 8x^2 - 15x + 4$ by $(3x-1)$:

$\frac{3x^3 + 8x^2 - 15x + 4}{3x - 1} = x^2 + 3x - 4$

So $3x^3 + 8x^2 - 15x + 4 = (3x - 1)(x^2 + 3x - 4)$

Now solve $x^2 + 3x - 4 = 0$. This factors as $(x+4)(x-1) = 0$. So $x = -4$ or $x = 1$.

The roots are $x = \frac{1}{3}, x = -4, x = 1$.

3. Final Answer

(i) $30 + 11\sqrt{6}$

(ii) $f'(x) = 6x$

(iii) $x = \frac{1}{3}, -4, 1$