Showing That a Polynomial is Divisible By 12

Suppose we want to show that  
\[4n^3+6n^2+14n\]
  is divisible by 12 for every integer  
\[n\]
. We can do this by looking at the remainders of  
\[, \]
  on division by 6 - the factor 2 means that  
\[n(2n^2+3n+7)\]
  needs to be divisible by 6.
\[n\]
 Remainder of  
\[n\]
  on Division By 6		 Remainder of  
\[2n^2+3n+7\]
  After Division By 6		 Remainder of  
\[n(2n^2+3n+7)\]
  After Division By 6
\[6k+1\]
 1 12 0
\[6k+2\]
 2 21/6=3 remainder 3 0
\[6k+3\]
 3 34/6=5 remainder 4 0
\[6k+4\]
 4 51/6=8 remainder 3 0
\[6k+5\]
 5 72/6=12 remainder 0 0
\[6k+6\]
 0 97 0

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