1. (a) Find the remainder when $x^3 - 2x^2 - 4x + 8$ is divided by (i) $x - 3$, (ii) $x + 2$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 2

Similarly, for $x + 2$, we evaluate:

$f(-2) = (-2)^3 - 2(-2)^2 - 4(-2) + 8$
$= -8 - 8 + 8 + 8$
$= 0$
Therefore, the remainder when $x^3 - 2x^2 - 4x + 8$ is divided by $x + 2$ is 0.

Since we found that $x + 2$ is a factor (with a remainder of 0), we can factor the polynomial as:

$x^3 - 2x^2 - 4x + 8 = (x + 2)(Ax^2 + Bx + C)$

To find $A$, $B$, and $C$, we can perform polynomial long division, or by comparing coefficients. After simplification, we get:

$(x + 2)(x^2 - 4) = (x + 2)(x - 2)(x + 2)$
Thus, we set:

$x + 2 = 0$
$x - 2 = 0$

Finding solutions: $x = -2, ext{ multiplicity 2}, x = 2$ Therefore, the solutions to the equation $x^3 - 2x^2 - 4x + 8 = 0$ are $x = -2$ (double root) and $x = 2$.