document.write( "Question 113774: find all real and imaginary roots\r
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Algebra.Com's Answer #82806 by Edwin McCravy(20056) $\"\"$ $\"About$
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document.write( "find all real and imaginary roots\r\n" );
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document.write( "Possible rational roots, if it has any,\r\n" );
document.write( "are ą1,ą2,ą4\r\n" );
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document.write( "DesCartes rule of signs tells us that\r\n" );
document.write( "there are no positive roots, so that\r\n" );
document.write( "narrows down the possible rational\r\n" );
document.write( "roots to -1, -2, and -4\r\n" );
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document.write( "Try , or \r\n" );
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document.write( "So we try dividing synthetically \r\n" );
document.write( "by . But we must first put in a\r\n" );
document.write( "zero term since the original\r\n" );
document.write( "equation does not contain a term\r\n" );
document.write( "in x.  So we rewrite the original\r\n" );
document.write( "equation, showing all coefficients: \r\n" );
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document.write( "-1|1  3  1  0  4\r\n" );
document.write( "  |  -1 -2  1 -1\r\n" );
document.write( "   1  2 -1  1  3\r\n" );
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document.write( "So we see that we don't get 0 as a\r\n" );
document.write( "remainder, the bottom right number,\r\n" );
document.write( "but rather 3, so -1 is not a root.\r\n" );
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document.write( "So we try  to see if it is a root.\r\n" );
document.write( "Dividing synthetically by  \r\n" );
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document.write( "-2|1  3  1  0  4\r\n" );
document.write( "  |  -2 -2  2 -4\r\n" );
document.write( "   1  1 -1  2  0\r\n" );
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document.write( "So  is a root, since we get 0\r\n" );
document.write( "as a remainder.  The 4 numbers to the\r\n" );
document.write( "left of the zero represent the quotient\r\n" );
document.write( "when we divided by , so we have now\r\n" );
document.write( "factored the polynomial equation as\r\n" );
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document.write( "So we now try to find a rational root of\r\n" );
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document.write( "DesCartes rule of signs tells us this\r\n" );
document.write( "has possible roots ą1 and ą2.  But since\r\n" );
document.write( "the original equation has no positive roots\r\n" );
document.write( "this can only have rational roots -1 and -2.\r\n" );
document.write( "We know -1 is not a root because it was not\r\n" );
document.write( "a root of the original equation, so we can\r\n" );
document.write( "only try  again as a root of multiplicity\r\n" );
document.write( "2.   is equivalent to . so we\r\n" );
document.write( "divide the new polynomial also by :\r\n" );
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document.write( "-2| 1  1 -1  2\r\n" );
document.write( "  |   -2  2 -2\r\n" );
document.write( "    1 -1  1  0\r\n" );
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document.write( "Yes, we get 0 as a remainder, so  is\r\n" );
document.write( "a root of multiplicity 2. So now our\r\n" );
document.write( "factorization of the original polynomial\r\n" );
document.write( "is now:\r\n" );
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document.write( "We set each factor = 0,\r\n" );
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document.write( " gives root \r\n" );
document.write( " gives root , again\r\n" );
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document.write( "This does not factor, so must be solved by the \r\n" );
document.write( "quadratic formula:\r\n" );
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document.write( "Its roots are \r\n" );
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document.write( "or, simplifying:\r\n" );
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document.write( "So there are four roots, counting multiplicities\r\n" );
document.write( "of roots as though they were separate roots:\r\n" );
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document.write( ", , , \r\n" );
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document.write( "Edwin
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