Answer: \(x=-3,\frac12,4\), giving \(A(-3,-23)\), \(B\left(\frac12,12\right)\), \(C(4,47)\). The midpoint of \(AC\) is \(B\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a) To show that \(x+3\) is a factor, substitute \(x=-3\) into the expression:

\(-12+23(-3)+3(-3)^2-2(-3)^3.\)

This is

\(-12-69+27+54=0.\)

Therefore \(x+3\) is a factor.

(b) At the intersections,

\(-5+33x+3x^2-2x^3=10x+7.\)

Rearranging gives

\(-12+23x+3x^2-2x^3=0.\)

Using part (a), factorise this cubic:

\(-12+23x+3x^2-2x^3=-(x+3)(2x-1)(x-4).\)

Hence the \(x\)-coordinates of the three intersection points are

\(-3,\quad \frac12,\quad 4.\)

The corresponding \(y\)-values can be found from the line \(y=10x+7\):

\(x=-3:\quad y=-30+7=-23,\)

\(x=\frac12:\quad y=5+7=12,\)

\(x=4:\quad y=40+7=47.\)

So

\(A(-3,-23),\quad B\left(\frac12,12\right),\quad C(4,47).\)

The midpoint of \(AC\) is

\(\left(\frac{-3+4}{2},\frac{-23+47}{2}\right) =\left(\frac12,12\right).\)

This is exactly the point \(B\), so \(B\) is the midpoint of \(AC\).