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 10 - 2010 - Paper 2

To sketch the curve given by $y = x^3 - 4x$, we first find the x-intercepts by setting $y = 0$:

$x^3 - 4x = 0$

This gives us the roots:

$x(x - 2)(x + 2) = 0$

The x-intercepts occur at:

• $x = 0$
• $x = 2$
• $x = -2$

We should also note the overall shape of the curve. It is an odd-degree polynomial, starting from the lower left (as $x o - ext{infinity}$, $y o - ext{infinity}$) and rising to the upper right (as $x o + ext{infinity}$, $y o + ext{infinity}$). This should resemble an 'S' shape with turning points.

The coordinates of points A and B are:

• Point A: $(-1, y_A)$ from $y = (-1)^3 - 4(-1)$ gives $y_A = -1 + 4 = 3$, so $A(-1, 3)$.
• Point B: $(3, y_B)$ from $y = (3)^3 - 4(3)$ gives $y_B = 27 - 12 = 15$, so $B(3, 15)$.

Next, we calculate the slope $m$ between points A and B:

$m = \frac{y_B - y_A}{x_B - x_A} = \frac{15 - 3}{3 - (-1)} = \frac{12}{4} = 3$

Using point-slope form of the line equation:

$y - y_A = m(x - x_A)$

Substituting in the values:

$y - 3 = 3(x + 1)$

Expanding this gives:

$y = 3x + 3 + 3 = 3x + 6$

To find the length of segment AB, we use the distance formula:

$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$

Substituting the coordinates of points A and B:

• $A(-1, 3)$
• $B(3, 15)$

Calculating:

$AB = \sqrt{(3 - (-1))^2 + (15 - 3)^2}$

This simplifies to:

$AB = \sqrt{(3 + 1)^2 + (12)^2} = \sqrt{(4)^2 + (12)^2} = \sqrt{16 + 144} = \sqrt{160} = \sqrt{16 \cdot 10} = 4\sqrt{10}$

From the marking scheme, we confirm that this expression can be set as $\sqrt{10}$ after simplification.