document.write( "Question 638383: solve y^2(y+1)=8y+12 \n" ); document.write( "
Algebra.Com's Answer #402648 by Edwin McCravy(20064)\"\" \"About 
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document.write( "        y²(y + 1) = 8y + 12\r\n" );
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document.write( "          y³ + y² = 8y + 12\r\n" );
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document.write( "y³ + y² - 8y - 12 = 0\r\n" );
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document.write( "Candidates for rational zeros are ± the divisors of 12\r\n" );
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document.write( "±1, ±2, ±3, ±4, ±6, ±12\r\n" );
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document.write( "Try 1\r\n" );
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document.write( " 1|1 1 -8 -12\r\n" );
document.write( "  |  1  2  -6 \r\n" );
document.write( "   1 2 -6 -18\r\n" );
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document.write( "No, that doesn't give a remainder of 0,\r\n" );
document.write( "so 1 is not a solution.\r\n" );
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document.write( "Try 2\r\n" );
document.write( " 2|1 1 -8 -12\r\n" );
document.write( "  |  2  6  -4 \r\n" );
document.write( "   1 3 -2 -16\r\n" );
document.write( "No, that doesn't give a remainder of 0,\r\n" );
document.write( "so 2 is not a solution.\r\n" );
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document.write( "Try 3\r\n" );
document.write( " 3|1 1 -8 -12\r\n" );
document.write( "  |  3 12  12 \r\n" );
document.write( "   1 4  4   0\r\n" );
document.write( "Yes, that gives a remainder of 0,\r\n" );
document.write( "so 3 is a solution.\r\n" );
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document.write( "And now we have factored the left\r\n" );
document.write( "side of\r\n" );
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document.write( "    y³ + y² - 8y - 12 = 0   as\r\n" );
document.write( " (y - 3)(y² + 4y + 4) = 0\r\n" );
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document.write( "The quadratic factor will also factor\r\n" );
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document.write( "(y - 3)(y + 2)(y + 2) = 0\r\n" );
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document.write( "Use the zero factor property:\r\n" );
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document.write( "   y = 3, y = -2, y = -2\r\n" );
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document.write( "So 3 is a solution of multiplicity 1\r\n" );
document.write( "and -2 is a solution of multiplicity 2.\r\n" );
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document.write( "Edwin

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