Question

b. \\(\frac{9x^2 - 4}{3x + 4}\\)

Explanation:

Step1: Factor the numerator

The numerator \(9x^2 - 4\) is a difference of squares, which can be factored as \((3x + 2)(3x - 2)\) since \(a^2 - b^2=(a + b)(a - b)\) where \(a = 3x\) and \(b = 2\). So we have \(\frac{(3x + 2)(3x - 2)}{3x + 4}\).

Step2: Check for common factors

The denominator is \(3x + 4\), and the numerator factors are \((3x + 2)\) and \((3x - 2)\). There are no common factors between the numerator and the denominator that can be canceled out. So the expression \(\frac{9x^2 - 4}{3x + 4}\) is already in its simplified form (or we can say that it cannot be simplified further by canceling common factors) and is equal to \(\frac{(3x + 2)(3x - 2)}{3x + 4}\). If we assume there was a typo and the denominator was \(3x + 2\) (a common type of problem), then we could cancel \((3x + 2)\) and get \(3x - 2\), but with the given denominator \(3x + 4\), we can't simplify by canceling.

Answer:

\(\frac{(3x + 2)(3x - 2)}{3x + 4}\) (or if denominator was a typo and should be \(3x + 2\), then \(3x - 2\))