document.write( "Question 594307: simplify fully
\n" ); document.write( "n!-2(n-2)! / (n-2)(n-2)! \n" ); document.write( "
Algebra.Com's Answer #376688 by mamiya(56)\"\" \"About 
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n!-2(n-2)! / (n-2)(n-2)!\r
\n" ); document.write( "\n" ); document.write( "n!-2(n-2)!/ (n-2)(n-2)! = (n(n-1)(n-2)! -2(n-2)! ) / ( (n-2)(n-2)!)
\n" ); document.write( " = (n(n-1) -2 )/ (n-2)
\n" ); document.write( " = (n^2 -n -2)/ (n-2)\r
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\n" ); document.write( "\n" ); document.write( " At this point, we have to look for the factors of n^2-n-2, if it exists\r
\n" ); document.write( "\n" ); document.write( "Using the quadratic formula, we get
\n" ); document.write( " n^2-n-2=0 , --> n= (1- sqrt( 1-4(-2)) )/2 or n = (1+sqrt( 1-4(-2))) /2
\n" ); document.write( " so n= -1 or n=2\r
\n" ); document.write( "\n" ); document.write( "this means the factorization of n^2-n-2 is (n-2)(n+1)
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\n" ); document.write( "\n" ); document.write( " so, n!-2(n-2)!/(n-2)(n-2)! = (n+1)(n-2)/ (n-2)
\n" ); document.write( " = n+1\r
\n" ); document.write( "\n" ); document.write( "so the complete simplification of n!-2(n-2)!/(n-2)(n-2)! is n+1\r
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