document.write( "Question 113886: Is there any one that knows how to work this problem?
\n" ); document.write( "Find the zeros of f.
\n" ); document.write( "f(x)= -x^(2)e^(-x)+ 2xe^(-x) \n" ); document.write( "

Algebra.Com's Answer #82869 by Edwin McCravy(20056) $\"\"$ $\"About$

You can put this solution on YOUR website!

\r\n" );
document.write( "Is there any one that knows how to work this problem?\r\n" );
document.write( "Find the zeros of f.\r\n" );
document.write( "\r\n" );
document.write( "To find the zeros substitute 0 for f(x) and solve for x.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Get 0 on the right:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Factor out \r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now we use the zero product principle.\r\n" );
document.write( "There are three factors: ,, and \r\n" );
document.write( "\r\n" );
document.write( "Setting x = 0 gives one solution, namely x = 0\r\n" );
document.write( "\r\n" );
document.write( "Setting  = 0 gives no solution, since  is never 0.\r\n" );
document.write( "\r\n" );
document.write( "Setting x-2 = 0 gives a second solution, namely x = 2\r\n" );
document.write( "\r\n" );
document.write( "So there are two zeros,  0 and 2.   \r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( "

\n" );