$f(x)$ is such that $f''(x) = 18x - 4$. Given that $f(0) = 1$ and $f(2) = 9$,

1. Integrate $f''(x)$ to find $f'(x)$

$f'(x) = \int (18x - 4) \, dx = 9x^2 - 4x + c$

2. Integrate $f'(x)$ to find the general form of $f(x)$

$f(x) = \int (9x^2 - 4x + c) \, dx = 3x^3 - 2x^2 + cx + d$

3. Use the condition $f(0) = 1$ to find $d$

$f(0) = 3(0)^3 - 2(0)^2 + c(0) + d = 1 \implies d = 1$
So, $f(x) = 3x^3 - 2x^2 + cx + 1$

4. Use the condition $f(2) = 9$ to find $c$ and state the final function

$f(2) = 3(2)^3 - 2(2)^2 + c(2) + 1 = 9$
$24 - 8 + 2c + 1 = 9$
$17 + 2c = 9$
$2c = -8 \implies c = -4$
Therefore, $f(x) = 3x^3 - 2x^2 - 4x + 1$

1. State the function $f(x)$ found in part (a)

From part (a), $f(x) = 3x^3 - 2x^2 - 4x + 1$.

2. Evaluate $f(-1)$ to test if $x+1$ is a factor

$f(-1) = 3(-1)^3 - 2(-1)^2 - 4(-1) + 1$
$f(-1) = -3 - 2 + 4 + 1$
$f(-1) = 0$

3. Conclude using the Factor Theorem

Since $f(-1) = 0$, by the Factor Theorem, $x + 1$ is a factor of $f(x)$.

1. State the polynomial $f(x)$ from part (a)

From part (a), $f(x) = 3x^3 - 2x^2 - 4x + 1$.

2. Factorise $f(x)$ using the factor $x+1$ from part (b)

Since $x+1$ is a factor, we can write $f(x) = (x+1)(ax^2+bx+c)$.
By comparing coefficients or long division:
$3x^3 - 2x^2 - 4x + 1 = (x+1)(3x^2 - 5x + 1)$

3. Solve the quadratic factor equal to zero

Set $3x^2 - 5x + 1 = 0$.
Using the quadratic formula:
$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(1)}}{2(3)}$
$x = \frac{5 \pm \sqrt{25 - 12}}{6}$
$x = \frac{5 \pm \sqrt{13}}{6}$

4. State all solutions to $f(x) = 0$

The solutions are $x = -1$ and $x = \frac{5 \pm \sqrt{13}}{6}$.