Question

q3 simplify the complex fraction completely. \\(\frac{\frac{\frac{x - 4}{2x - 1}}{\frac{x^2 - 16}{2x - 1}}}{\frac{1}{x + 4}}\\) (with multiple choice options: \\(\frac{1}{x + 4}\\), \\(x - 4\\), and another polynomial fraction)

Explanation:

Step1: Rewrite division as multiplication

A complex fraction \(\frac{\frac{a}{b}}{\frac{c}{d}}\) is equivalent to \(\frac{a}{b} \div \frac{c}{d}=\frac{a}{b}\times\frac{d}{c}\). So our complex fraction \(\frac{\frac{x - 4}{2x - 1}}{\frac{x^{2}-16}{2x - 1}}\) becomes \(\frac{x - 4}{2x - 1}\times\frac{2x - 1}{x^{2}-16}\).

Step2: Factor the difference of squares

Recall that \(a^{2}-b^{2}=(a - b)(a + b)\). Here, \(x^{2}-16=x^{2}-4^{2}=(x - 4)(x + 4)\). So we substitute this into our expression: \(\frac{x - 4}{2x - 1}\times\frac{2x - 1}{(x - 4)(x + 4)}\).

Step3: Cancel common factors

We can cancel out the common factors \((x - 4)\) and \((2x - 1)\) from the numerator and the denominator. After canceling, we are left with \(\frac{1}{x + 4}\).

Answer:

\(\frac{1}{x + 4}\) (corresponding to the option \(\boldsymbol{\frac{1}{x + 4}}\))