(8) \( \frac{2^{2 x}-9}{2^{x}+3} \)

Sure! Let’s dive a bit deeper into this expression. First, we can rewrite the numerator \(2^{2x}\) as \((2^x)^2\). So now the expression looks like \(\frac{(2^x)^2 - 9}{2^x + 3}\). This resembles the difference of squares, which is \(a^2 - b^2 = (a - b)(a + b)\), where in our case \(a = 2^x\) and \(b = 3\). This means we can factor the numerator as \((2^x - 3)(2^x + 3)\). Next, we simplify! The denominator is \(2^x + 3\), which we can cancel with the factor in the numerator. This leaves us with the simplified expression: \(2^x - 3\), assuming \(2^x \neq -3\) (which is always true since \(2^x\) is positive). Voilà! Now you've got a much simpler expression to work with!