If \(f(x)=x^2+4x+4\) and \(g(x)=x+2\), what is the value of \(\left(\frac{f}{g}\right)(x)\) for \(x\ne -2\)?

Correct answer: \(x+2\)

\(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2+4x+4}{x+2}\). Factoring the numerator as \((x+2)^2\) gives \(\frac{(x+2)^2}{x+2}=x+2\), provided \(x\ne -2\). Option C is not correct because it equals the quotient only for a particular value, not for all allowed values of \(x\). Exam tip: factor the numerator first and exclude any value that makes the original denominator zero.