Question 3.ShareReportWhat is the largest positive integer n such that n2+7n+12n2−n−12\frac{n^2 + 7n + 12}{n^2 - n - 12}n2−n−12n2+7n+12 is also a positive integer?A6B16C8D12Backspace7894561230.-Clear AllSubmit Previous QuestionRate this SolutionNext Question Question ExplanationText Explanation n2+3n+4n+12n2−4n+3n−12\ \frac{\ n^2+3n+4n+12}{n^2-4n+3n-12} n2−4n+3n−12 n2+3n+4n+12= n(n+3)+4(n+3)n(n−4)+3(n−4)\ \frac{\ n^{ }\left(n+3\right)+4\left(n+3\right)}{n^{ }\left(n-4\right)+3\left(n-4\right)} n(n−4)+3(n−4) n(n+3)+4(n+3)= ( n+4)(n+3)(n−4)(n+3)\ \frac{\left(\ n+4\right)\left(n+3\right)}{\left(n-4\right)\left(n+3\right)} (n−4)(n+3)( n+4)(n+3)= ( n+4)(n−4)\ \frac{\left(\ n+4\right)}{\left(n-4\right)} (n−4)( n+4)= ( n−4)+8(n−4)\ \frac{\left(\ n-4\right)+8}{\left(n-4\right)} (n−4)( n−4)+8= 1+8(n−4)\ 1+\frac{8}{\left(n-4\right)} 1+(n−4)8 which will be maximum when n-4 =8n=12D is the correct answer.Video ExplanationNo video explanation yet — we're on it and uploading soon!
