It is given that $f(x) = ax^3 + 3x^2 + bx + 1$ where $a$ and $b$ are constants.

1. Find the values of $a$ and $b$ from part (i)

From (i), $f(-1) = 0 \implies -a + 3 - b + 1 = 0 \implies a + b = 4$.
$f(0.5) = 3 \implies a(0.125) + 3(0.25) + 0.5b + 1 = 3 \implies 0.125a + 0.5b = 1.25 \implies a + 4b = 10$.
Solving simultaneously: $3b = 6 \implies b = 2$, $a = 2$.
So $f(x) = 2x^3 + 3x^2 + 2x + 1$.

2. Factorise $f(x)$ using the known factor $(x+1)$

Since $f(-1) = 0$, $(x+1)$ is a factor.
$2x^3 + 3x^2 + 2x + 1 = (x+1)(2x^2 + cx + 1)$
Comparing coefficients of $x^2$: $c + 2 = 3 \implies c = 1$.
So $f(x) = (x+1)(2x^2 + x + 1)$.

3. Analyse the roots of the quadratic factor

For the equation $f(x) = 0$, either $x+1 = 0$ or $2x^2 + x + 1 = 0$.
For $2x^2 + x + 1 = 0$, the discriminant is $b^2 - 4ac = 1^2 - 4(2)(1) = 1 - 8 = -7$.
Since the discriminant is less than 0, the quadratic equation $2x^2 + x + 1 = 0$ has no real roots.

4. Conclude the number of real roots and state the root

Therefore, the only real root comes from $x+1 = 0$, which gives $x = -1$.

1. State the full polynomial $f(x)$ using the values of $a$ and $b$ found in part (i)

From part (i), $a = 2$ and $b = 2$.
$f(x) = 2x^3 + 3x^2 + 2x + 1$

2. Apply the Remainder Theorem to find the remainder when divided by $x - 4$

By the Remainder Theorem, the remainder is $f(4)$:
$\text{Remainder} = f(4)$
$f(4) = 2(4)^3 + 3(4)^2 + 2(4) + 1$
$f(4) = 2(64) + 3(16) + 8 + 1$
$f(4) = 128 + 48 + 9 = 185$