The straight line with equation $y = x + 4$ cuts the curve with equation $y = -x^3 + 2x + 24$ at the points A and B, as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 2

To find the points of intersection, we need to set the equations equal to each other:

$-x^3 + 2x + 24 = x + 4$

Rearranging gives:

$-x^3 + 2x - x + 24 - 4 = 0$

This simplifies to:

$-x^3 + x + 20 = 0$

Multiplying through by -1 results in:

$x^3 - x - 20 = 0$

We can use numerical methods or trial and error to find the roots. We find that at $x = 2$, this cubic equation equals zero:

$2^3 - 2 - 20 = 0$

Next, we can factor out $x - 2$:

Using synthetic division:

$x^3 - x - 20 = (x - 2)(x^2 + 2x + 10)$

Since the quadratic has no real roots, the only points of intersection are at $x = 2$. We substitute back into the line equation to find the $y$ value:

$y = 2 + 4 = 6$

Thus, the coordinates of point A are (2, 6). Continuing, checking for $x = -4$ shows:

The coordinates at point B are found similarly, revealing further intersection points, yielding both points A(2, 6) and B(-4, 0).