Question

what is the difference in simplest form?

n^2+3n+2/n^2+6b+8 - 2n/n+4

Answer 1

\[\frac{ n^2+3n+2 }{ (n+4)(n+2) }-\frac{ 2n }{ n+4 }=\frac{ n^2+3n+2 }{ (n+4)(n+2) }-\frac{ 2n(n+2) }{(n+4)(n+2) }\]\[=\frac{ n^2+3n+2 }{ (n+4)(n+2) }-\frac{ 2n^2+4n }{(n+4)(n+2) }=\frac{ (n^2+3n+2)-(2n^2+4n) }{ (n+4)(n+2) }\]\[=\frac{ -n^2-n+2}{ (n+4)(n+2) }=\frac{ -(n+2)(n-1)}{ (n+4)(n+2) }\]\[=-\frac{ n-1}{ n+4 }\]
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Answer 2

1-n?

Answer 3

No, after you subtract, the final answer is:\[-\frac{ n-1 }{ n+4 }\]
@OSMiUM_MONkEY

Answer 4

Get it? And btw, don't forget the medal =D

Answer 5

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