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 3 - 2008 - Paper 2

Now, we need to find the remainder when dividing by $x + 2$, which also uses the Remainder Theorem. We calculate $f(-2)$:

$f(-2) = (-2)^3 - 2(-2)^2 - 4(-2) + 8$ $= -8 - 8 + 8 + 8$ $= 0$

Therefore, the remainder when dividing by $x + 2$ is 0.

From part (ii), we found that $x + 2$ is a factor since the remainder is zero. Thus, we can factor the polynomial as:

$x^3 - 2x^2 - 4x + 8 = (x + 2)(Q(x))$

where $Q(x)$ is the quotient polynomial of degree 2. To find $Q(x)$, we can use polynomial long division, or directly factor:

Perform polynomial division:

$x^3 - 2x^2 - 4x + 8 \div (x + 2) \rightarrow Q(x) = x^2 - 4$

So, we have:

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

Next, we can factor $x^2 - 4$ as:

$(x - 2)(x + 2)$

Thus, the complete factorization is:

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

Finally, setting the equation to zero, we find:

1. $x + 2 = 0$ $$$$

ightarrow x = -2$$ 2. $x - 2 = 0$

ightarrow x = 2$$ Therefore, the solutions to the equation $x^3 - 2x^2 - 4x + 8 = 0$ are $x = -2$ (with multiplicity 2) and $x = 2$.