Question

final exam - requires respondus lockdown browser + webcam
started: dec 10 at 8:58pm
quiz instructions
question 32
1 pts
multiply. write the answer in lowest terms.
\\(\frac{x^2 + 16x + 64}{x^2 - 36} cdot \frac{x - 6}{x + 8}\\)
\\(\bigcirc \frac{x + 8}{(x + 6)^2}\\)
\\(\bigcirc \frac{x + 8}{x + 6}\\)
\\(\bigcirc \frac{x + 8}{x - 6}\\)
\\(\bigcirc \frac{(x + 8)^2}{(x + 6)^2}\\)
\\(\bigcirc \frac{x - 8}{x + 6}\\)

Explanation:

Step1: Factor the expressions

Factor \(x^2 + 16x + 64\) (perfect square trinomial: \((a + b)^2=a^2 + 2ab + b^2\), here \(a = x\), \(b = 8\), so \(x^2+16x + 64=(x + 8)^2\)). Factor \(x^2-36\) (difference of squares: \(a^2 - b^2=(a + b)(a - b)\), here \(a=x\), \(b = 6\), so \(x^2-36=(x + 6)(x - 6)\)). The expression becomes \(\frac{(x + 8)^2}{(x + 6)(x - 6)}\cdot\frac{x - 6}{x + 8}\).

Step2: Cancel common factors

Cancel out the common factors \((x - 6)\) (from numerator and denominator) and one \((x + 8)\) (from numerator and denominator). After canceling, we have \(\frac{x + 8}{x + 6}\).

Answer:

\(\boldsymbol{\frac{x + 8}{x + 6}}\) (corresponding to the option with this expression, e.g., if the second option is \(\frac{x + 8}{x + 6}\), then the answer is the option with that text, like "B. \(\frac{x + 8}{x + 6}\)" depending on the option numbering)