Show that $$\frac{6x^3}{9x^2 - 144} \div \frac{2x^4}{3(x - 4)}$$ can be written in the form $$\frac{1}{x(x + r)}$$ where r is an integer. - Edexcel - GCSE Maths - Question 23 - 2021 - Paper 3

Next, we factor the denominator. The expression $9x^2 - 144$ can be factored as follows:

$9x^2 - 144 = 9(x^2 - 16) = 9(x - 4)(x + 4)$

Substituting this back into our expression gives us:

$\frac{6x^3 \times 3(x - 4)}{9 \times (x - 4)(x + 4) \times 2x^4}$

Now, we can simplify the expression. Cancel the common factors of $(x - 4)$:

$\frac{6x^3 \times 3}{9 \times 2x^4(x + 4)} = \frac{18x^3}{18x^4(x + 4)}$

This simplifies to:

$\frac{1}{x(x + 4)}$

Thus, we can conclude that $r = 4$, which is indeed an integer.