The curve C has equation y = (x + 3)(x - 1)^2 - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 2

Starting from the original equation:

$y = (x + 3)(x - 1)^2$

First, expand $(x - 1)^2$:

$(x - 1)^2 = x^2 - 2x + 1$

Substituting this back:

$y = (x + 3)(x^2 - 2x + 1)$

Now, distribute:

$y = x(x^2 - 2x + 1) + 3(x^2 - 2x + 1)$

$y = x^3 - 2x^2 + x + 3x^2 - 6x + 3$

Combining like terms gives:

$y = x^3 + (3 - 2)x^2 + (-6 + 1)x + 3$

$y = x^3 + x^2 - 5x + 3$

Thus, comparing this with the required form, we see that $k = 3$. Therefore, the value of k is 3.

To find the points where the gradient of the tangent to C is equal to 3, we first need to determine the derivative of y:

$y = x^3 + x^2 - 5x + 3$

Calculating the derivative:

$\frac{dy}{dx} = 3x^2 + 2x - 5$

Setting the derivative equal to 3 gives:

$3x^2 + 2x - 5 = 3$

Simplifying this, we have:

$3x^2 + 2x - 8 = 0$

Now, we can use the quadratic formula to find the x-coordinates:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 3$, $b = 2$, and $c = -8$:

$x = \frac{-2 \pm \sqrt{(2)^2 - 4(3)(-8)}}{2(3)}$

Calculating the discriminant:

$= 4 + 96 = 100$

Substituting back:

$x = \frac{-2 \pm 10}{6}$

This gives two solutions:

1. $x = \frac{8}{6} = \frac{4}{3}$
2. $x = \frac{-12}{6} = -2$

Thus, the x-coordinates of the two points where the gradient of the tangent is equal to 3 are $\frac{4}{3}$ and $-2$.