The curve C with equation $y = f(x)$ passes through the point $(5, 65)$.\nGiven that $f'(x) = 6x^2 - 10x - 12$,\n(a) use integration to find $f(x)$.\n\n(b) Hence show that $f(x) = x(2x + 3)(x - 4)$.\n\n(c) In the space provided on page 17, sketch C, showing the coordinates of the points where C crosses the $x$-axis. - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 1

To show that $f(x) = 2x^3 - 5x^2 - 12x$ can be factored as $x(2x + 3)(x - 4)$, we first express $f(x)$ in factored form:

1. Factor out $x$: $f(x) = x(2x^2 - 5x - 12)$

2. Now, we factor the quadratic $2x^2 - 5x - 12$. Using the quadratic formula, we can find its roots:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(-12)}}{2(2)} = \frac{5 \pm \sqrt{25 + 96}}{4} = \frac{5 \pm 11}{4}$

   This results in: $x = 4 \quad \text{and} \quad x = -1.5$

3. Therefore, the factored form is: $f(x) = x(2(x - 4)(x + 1.5))$

   Hence, we can rewrite this as: $f(x) = x(2x + 3)(x - 4)$.

To determine where the curve $C$ crosses the $x$-axis, we set $f(x) = 0$:

1. We have already factored $f(x)$ as: $f(x) = x(2x + 3)(x - 4)$

2. Setting this equal to zero gives:

   • $x = 0$
   • $2x + 3 = 0 \Rightarrow x = -1.5$
   • $x - 4 = 0 \Rightarrow x = 4$

Thus, the points where the curve crosses the $x$-axis are:

• $(0, 0)$
• $(-1.5, 0)$
• $(4, 0)$.

For the sketch, plot these points on the $x$-axis and ensure the curve has the correct shape, passing through these points.