document.write( "Question 63890: Need help....please.\r
\n" ); document.write( "\n" ); document.write( "Use Descartes Rule to determine how many positive and negative zeros.
\n" ); document.write( "Do not need to find the zeros.\r
\n" ); document.write( "\n" ); document.write( "f(x)= -6x^5+x^4+5x^3+x+1 \n" ); document.write( "
Algebra.Com's Answer #44536 by Edwin McCravy(20086)\"\" \"About 
You can put this solution on YOUR website!
Need help....please. \r\n" );
document.write( "Use Descartes Rule to determine how many positive and negative zeros.\r\n" );
document.write( "Do not need to find the zeros. \r\n" );
document.write( "f(x)= -6x5 + x4 + 5x3 + x + 1\r\n" );
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document.write( "Rule:\r\n" );
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document.write( "Write the terms in descending order, then list the signs of the terms:\r\n" );
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document.write( "f(x)= -6x5 + x4 + 5x3 + x + 1\r\n" );
document.write( "      |    |    |     |   |\r\n" );
document.write( "      -    +    +     +   +\r\n" );
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document.write( "In going from the 1st sign - to the 2nd sign +, that is 1 sign\r\n" );
document.write( "change.\r\n" );
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document.write( "In going from the 2nd sign + to the 3rd sign +, there is no\r\n" );
document.write( "sign change.  So there is still only one sign change.\r\n" );
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document.write( "In going from the 3rd sign + to the 4th sign +, there is no\r\n" );
document.write( "sign change.  So there is still only one sign change.\r\n" );
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document.write( "In going from the 4th sign + to the 5th sign +, there is no\r\n" );
document.write( "sign change.  So there is still only one sign change.\r\n" );
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document.write( "So there is one sign change, so there is just one positive\r\n" );
document.write( "zero.  \r\n" );
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document.write( "*If there had been more than 1 sign change, the number of\r\n" );
document.write( "positive zeros might have been reduced by a multiple of 2,\r\n" );
document.write( "but we can't reduce 1 by a multiple of 2, so there must be\r\n" );
document.write( "exactly one positive zero.\r\n" );
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document.write( "Next to find the number of negative zeros we must find f(-x).\r\n" );
document.write( "The positive zeros of f(-x) will be the zeros of f(x) with\r\n" );
document.write( "their sign changed, so we just need to do the same thing with\r\n" );
document.write( "f(-x) as we did above. \r\n" );
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document.write( "f(-x)= -6(-x)5 + (-x)4 + 5(-x)3 + (-x) + 1\r\n" );
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document.write( "f(-x) = +6x5 + x4 - 5x3 - x + 1\r\n" );
document.write( "        |    |    |     |   |\r\n" );
document.write( "        +    +    -     -   +\r\n" );
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document.write( "In going from the 1st sign + to the 2nd sign +, there is no\r\n" );
document.write( "sign change.\r\n" );
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document.write( "In going from the 2nd sign + to the 3rd sign +, there is one\r\n" );
document.write( "sign change.  \r\n" );
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document.write( "In going from the 3rd sign - to the 4th sign -, there is no\r\n" );
document.write( "sign change.  So there is still only one sign change.\r\n" );
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document.write( "In going from the 4th sign - to the 5th sign +, there is one\r\n" );
document.write( "sign change.  So we end up with two sign changes.\r\n" );
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document.write( "Since there are two sign change, there will either two of\r\n" );
document.write( "0 negative zeros.  However the rule can't tell us whether it\r\n" );
document.write( "is 2 negative zeros or 0 negative zeros.\r\n" );
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document.write( "Note that if the signs of some polynomial were:\r\n" );
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document.write( " -  +  -  +  -  +  -  -\r\n" );
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document.write( "There would be 6 sign changes, so there would be\r\n" );
document.write( "either 6 or 4 or 2 or 0 positive zeros.\r\n" );
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document.write( "If the signs of some polynomial were\r\n" );
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document.write( " +  -  +  -  +  -  +  -\r\n" );
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document.write( "There would be 7 sign changes, so there would be\r\n" );
document.write( "either 7 or 5 or 3 or 1 positive zeros.  \r\n" );
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document.write( "Note that if there are an odd number of sign changes,\r\n" );
document.write( "there must be at least one positive zero.  However if\r\n" );
document.write( "there are an even number of sign changes there may \r\n" );
document.write( "not be any positive zeros at all.  \r\n" );
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document.write( "Edwin

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