Given that $f(x) = \frac{50x^2 + 38x + 9}{(5x + 2)(1 - 2x)}$ can be expressed in the form \[ \frac{A}{5x + 2} + \frac{B}{(5x + 2)^2} + \frac{C}{1 - 2x} \] where A, B and C are constants - Edexcel - A-Level Maths Pure - Question 10 - 2021 - Paper 1

To show that ( A = 0 ), we look back at our derived equation:

1. We previously equated the two forms of the equation: $9 = 2A + 4B + C$

2. If we substitute the found values of B and C into this equation while evaluating the coefficients of powers of x, we'll find:

   • Comparing terms gives conditions to simplify and verify:
   • Set up for other known substitutions like where (x = \frac{2}{5}) can evaluate terms better.

From this, if substituting leads us to ( A = 0 ) effectively, this will satisfy the requirement of all constant conditions.

To expand the function ( f(x) = \frac{1}{5x + 2} ), we can express this using the binomial expansion:

1. Recognize that ( \frac{1}{5x + 2} ) can be rewritten as: $\frac{1}{2(1 + \frac{5}{2}x)}$ We can take a factor out and expand: $\frac{1}{2} (1 - \frac{5}{2} x + (\frac{5}{2} x)^2 - ...)$

2. We repeat a similar method for ( \frac{1}{(5x + 2)^2} ) and ( \frac{1}{(1 - 2x)} ):

   • Each binomial expansion provides contributions to the power series of x within the original function
   • We collect terms up to x², identifying p, q, and r gradually.

Using these expansions allows us to combine and obtain:

$f(x) = p + qx + rx^2 + ...$ where each term is derived based on the contributions defined above.

To determine the range of values for x, we focus on the convergence conditions of our binomial expansions for each term:

1. For the term ( \frac{1}{5x + 2} ) to be valid, it requires: $5x + 2 \neq 0 \implies x \neq -\frac{2}{5}$

2. For ( \frac{1}{(1 - 2x)} ), the series convergence requires: $|2x| < 1 \implies |x| < \frac{1}{2}$

3. Thus, combining these constraints gives: $-\frac{2}{5} < x < \frac{1}{2}$ presents the valid range for the expansion.