(a)

(x + x^{2} + x^{3} + x^{4})^{2} = x^{2} + 2 x^{3} + 3 x^{4} + 4 x^{5} + 3 x^{6} + 2 x^{7} + x^{8}

(b)

\begin{matrix} Scores & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ Frequencies & 1 & 2 & 3 & 4 & 3 & 2 & 1 \\ Relative frequencies & 0.0625 & 0.125 & 0.1875 & 0.25 & 0.1875 & 0.125 & 0.0625 \end{matrix}

(d) Here is where users can experiment with Mike's Probability Environment. (e) Possible answers 1, 3, 3, 5 and 1, 2, 2, 3 or 0, 2, 2, 4 and 2, 3, 3, 4 or 2, 4, 4, 6 and 0, 1, 1, 2

(f)The spinner-pairs in (e) correspond to the three factorisations below:

$(x + x^{2} + x^{3} + x^{4})^{2} = x^{2} (1 + x)^{2} (1 + x^{2})^{2} .$

$\begin{matrix} [x (1 + x^{2})^{2}] [x (1 + x)^{2}] & = [x + 2 x^{3} + x^{5}] [x + 2 x^{2} + x^{3}] & (1) \\ [(1 + x^{2})^{2}] [x^{2} (1 + x)^{2}] & = [1 + 2 x^{2} + x^{4}] [x^{2} + 2 x^{3} + x^{4}] & (2) \\ [x^{2} (1 + x^{2})^{2}] [(1 + x)^{2}] & = [x^{2} + 2 x^{4} + x^{6}] [1 + 2 x + x^{2}] . & (3) \end{matrix}$

(g) Also

$\begin{matrix} (x + x^{2} + x^{3} + x^{4})^{2} & = x^{2} (1 + x)^{2} (1 + x^{2})^{2} \\ = [x^{2} (1 + x)] [(1 + x) (1 + x^{2})^{2}] \\ = [x^{2} + x^{3}] [1 + x + 2 x^{2} + 2 x^{3} + x^{4} + x^{5}] \end{matrix}$

suggests a 2-spinner labelled 2, 3 and an 8-spinner labelled 0, 1, 2, 2, 3, 3, 4, 5.