9. (a) Factorise completely $$x^3 - 4x$$ (b) Sketch the curve C with equation $$y = x^3 - 4x$$, showing the coordinates of the points at which the curve meets the x-axis - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 2

To find the points where the curve meets the x-axis, set y to 0:

$0 = x^3 - 4x$ This gives: $x(x - 2)(x + 2) = 0$ The solutions are: $x = 0, x = 2, x = -2$

These points on the x-axis are (-2, 0), (0, 0), and (2, 0). The graph will have a shape resembling an cubic function, crossing the x-axis at these points, and it should also touch the origin.

Given points A (-1, f(-1)) and B (3, f(3)), we need to calculate the coordinates of function at these points:

First, calculate f(-1): $f(-1) = (-1)^3 - 4(-1) = -1 + 4 = 3$
So, A = (-1, 3).

Now, calculate f(3): $f(3) = (3)^3 - 4(3) = 27 - 12 = 15$
So, B = (3, 15).

The slope (m) of the line AB is: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{15 - 3}{3 - (-1)} = \frac{12}{4} = 3$

Using the point-slope formula, we have: $y - 3 = 3(x + 1)$ Simplifying gives the line equation: $y = 3x + 6$

The distance AB is calculated using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting A and B: $d = \sqrt{(3 - (-1))^2 + (15 - 3)^2}$ $= \sqrt{(3 + 1)^2 + (12)^2}$ $= \sqrt{(4)^2 + (12)^2} = \sqrt{16 + 144} = \sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10}$ So, the length of AB can be expressed as: $AB = 4\sqrt{10}$