document.write( "Question 288534: use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for the given function.
\n" ); document.write( "f(x)=x^5-1.7x^4-17.65x^3+3x^2+45x-15.411 \n" ); document.write( "
Algebra.Com's Answer #209150 by Edwin McCravy(20056) $\"\"$ $\"About$
You can put this solution on YOUR website!
\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The 1st term is , That has a positive sign.\r\n" );
document.write( "\r\n" );
document.write( "The 2nd term is .  That has a negative sign.  Therefore\r\n" );
document.write( "going from the positive 1st term to the negative 2nd term is a sign \r\n" );
document.write( "change.  So far there is 1 sign change.\r\n" );
document.write( "\r\n" );
document.write( "The 3rd term is .  That has a negative sign.  Therefore\r\n" );
document.write( "going from the negative 2nd term to the negative 3rd term is NOT a sign \r\n" );
document.write( "change.  So far there is still just 1 sign change.\r\n" );
document.write( "\r\n" );
document.write( "The 4th term is .  That has a positive sign.  Therefore\r\n" );
document.write( "going from the negative 3rd term to the positive 4th term is a sign \r\n" );
document.write( "change.  So far there are now 2 sign changes.\r\n" );
document.write( "\r\n" );
document.write( "The 5th term is .  That has a positive sign.  Therefore\r\n" );
document.write( "going from the positive 4th term to the positive 5th term is NOT a sign \r\n" );
document.write( "change.  So far there are still just 2 sign changes.\r\n" );
document.write( "\r\n" );
document.write( "The 6th term is .  That has a negative sign.  Therefore\r\n" );
document.write( "going from the positive 5th term to the negative 6th term is a sign \r\n" );
document.write( "change.  So far there are now 3 sign changes.\r\n" );
document.write( "\r\n" );
document.write( "We have reached the end of the polynomial.  It has 3 sign changes.\r\n" );
document.write( "\r\n" );
document.write( "So there could be 3 positive zeros.  There could also be 1 positive\r\n" );
document.write( "zero, because the number of positive zeros could either be the number\r\n" );
document.write( "of sign changes, or a multiple of 2 fewer positive zeros. \r\n" );
document.write( "\r\n" );
document.write( "So there are either 3 or 1 positive zero.\r\n" );
document.write( "\r\n" );
document.write( "Next substitute  for  and simplify:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The 1st term is , That has a negative sign.\r\n" );
document.write( "\r\n" );
document.write( "The 2nd term is .  That has a negative sign.  Therefore\r\n" );
document.write( "going from the negative 1st term to the negative 2nd term is NOT a sign \r\n" );
document.write( "change.  So far there are no sign changes.\r\n" );
document.write( "\r\n" );
document.write( "The 3rd term is .  That has a positive sign.  Therefore\r\n" );
document.write( "going from the negative 2nd term to the positive 3rd term is a sign \r\n" );
document.write( "change.  So far there is now just 1 sign change.\r\n" );
document.write( "\r\n" );
document.write( "The 4th term is .  That has a positive sign.  Therefore\r\n" );
document.write( "going from the positive 3rd term to the positive 4th term is NOT a sign \r\n" );
document.write( "change.  So far there is still just 1 sign change.\r\n" );
document.write( "\r\n" );
document.write( "The 5th term is .  That has a negative sign.  Therefore\r\n" );
document.write( "going from the positive 4th term to the negative 5th term is a sign \r\n" );
document.write( "change.  So far there are now 2 sign changes.\r\n" );
document.write( "\r\n" );
document.write( "The 6th term is .  That has a negative sign.  Therefore\r\n" );
document.write( "going from the negative 5th term to the negative 6th term is NOT a sign \r\n" );
document.write( "change.  So far there are still 2 sign changes.\r\n" );
document.write( "\r\n" );
document.write( "So there could be 2 negative zeros.  There could also be 0 negative\r\n" );
document.write( "zeros, because the number of negative zeros could either be the number\r\n" );
document.write( "of sign changes, or a multiple of 2 fewer negative zeros. \r\n" );
document.write( "\r\n" );
document.write( "So there are either 2 or 0 negative zeros.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( " \n" ); document.write( "
\n" );