document.write( "Question 701219: Factor completely. If the polynomial is prime, state this.\r
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Algebra.Com's Answer #432314 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "-2x²+5x+12\r\n" );
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document.write( "Since the x² term is negative, we will \r\n" );
document.write( "factor out -1:\r\n" );
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document.write( "-1(2x² - 5x - 12)\r\n" );
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document.write( "or just:\r\n" );
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document.write( "-(2x² - 5x - 12)\r\n" );
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document.write( "There is only one way to factor the 2x², as x and 2x\r\n" );
document.write( "So we write\r\n" );
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document.write( "-(x  )(2x  )\r\n" );
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document.write( "The ways to factor 12 are 12·1, 1·12, 6·2, 2·6, 4·3, 3·4\r\n" );
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document.write( "-(x 12)(2x 1)\r\n" );
document.write( "-(x 1)(2x 12)\r\n" );
document.write( "-(x 6)(2x 2)\r\n" );
document.write( "-(x 2)(2x 6)\r\n" );
document.write( "-(x 4)(2x 3)\r\n" );
document.write( "-(x 3)(2x  4)\r\n" );
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document.write( "There was no common factor to take out of the\r\n" );
document.write( " 3 terms in the parentheses, so we can\r\n" );
document.write( "eliminate any of those that have a common factor.\r\n" );
document.write( "So we have only these\r\n" );
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document.write( "-(x 12)(2x 1)\r\n" );
document.write( "-(x 4)(2x 3)\r\n" );
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document.write( "The only promising one is the second one\r\n" );
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document.write( "-(x 4)(2x 3)\r\n" );
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document.write( "To make the middle term come out as -5x,\r\n" );
document.write( "we have to put the signs in this way:\r\n" );
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document.write( "-(x-4)(2x+3)\r\n" );
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document.write( "Edwin
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