Question

perform the indicated operation, and simplify if possible. assume any factors you cancel are not zero. (dfrac{dfrac{c - 8}{c^2 - 16}}{1 - dfrac{12}{c + 4}} = dfrac{square}{square}) note: you can earn partial credit on this problem. you have attempted this problem 0 times. you have unlimited attempts remaining. (interface elements: maa, webwork, cal1johal, review p.ii, problem 24, previous problem, problem list, next problem, preview my answers, submit answers)

Explanation:

Step1: Simplify the denominator of the complex fraction

First, simplify \(1 - \frac{12}{c + 4}\). Find a common denominator, which is \(c + 4\). So we have \(\frac{c + 4}{c + 4}-\frac{12}{c + 4}=\frac{(c + 4)-12}{c + 4}=\frac{c - 8}{c + 4}\).

Step2: Simplify the numerator of the complex fraction

The numerator is \(\frac{c - 8}{c^{2}-16}\). Notice that \(c^{2}-16\) is a difference of squares, so \(c^{2}-16=(c - 4)(c + 4)\). So the numerator becomes \(\frac{c - 8}{(c - 4)(c + 4)}\).

Step3: Divide the numerator by the denominator

Now we have a complex fraction \(\frac{\frac{c - 8}{(c - 4)(c + 4)}}{\frac{c - 8}{c + 4}}\). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes \(\frac{c - 8}{(c - 4)(c + 4)}\times\frac{c + 4}{c - 8}\).

Step4: Cancel common factors

Cancel out the common factors \(c - 8\) and \(c + 4\) (assuming \(c
eq8\) and \(c
eq - 4\) and \(c
eq4\) as given). We are left with \(\frac{1}{c - 4}\).

Answer:

\(\frac{1}{c - 4}\)