document.write( "Question 331322: factor the polynomial completely x^3+8X^2+19X+12 \n" ); document.write( "

Algebra.Com's Answer #237461 by Edwin McCravy(20054) $\"\"$ $\"About$

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factor the polynomial completely $\"x%5E3%2B8x%5E2%2B19x%2B12\"$
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document.write( "All feasible rational zeros are  the factors of 12 \r\n" );
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document.write( " , , , , , ,   \r\n" );
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document.write( "Also there are no sign changes so there are no positive zeros.\r\n" );
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document.write( "So we begin by trying -1\r\n" );
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document.write( " -1| 1  8 19  12\r\n" );
document.write( "   |   -1 -7 -12\r\n" );
document.write( "     1  7 12   0\r\n" );
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document.write( "So x = -1 is a zero, so therefore (x + 1) is a factor.\r\n" );
document.write( "So now we have factored the original polynomial as\r\n" );
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document.write( "Now we factor the trinomial on the right and get:\r\n" );
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document.write( "That's it.\r\n" );
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document.write( "Edwin

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