document.write( "Question 1024496: Factorise (x-1)(x-2)(x+3)(x+4)+4 \n" ); document.write( "
Algebra.Com's Answer #639911 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Yes it is factorable.  The other tutor thought\r\n" );
document.write( "because it has no rational zeros (roots) that it\r\n" );
document.write( "is not factorable.  That is not the case.  A 4th\r\n" );
document.write( "degree polynomial could factor into the product \r\n" );
document.write( "of two quadratic trinomials both which have \r\n" );
document.write( "irrational zeros (roots).\r\n" );
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document.write( "Here is the correct approach:\r\n" );
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document.write( "(x-1)(x-2)(x+3)(x+4)+4\r\n" );
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document.write( "If we were going to multiply that out, we would\r\n" );
document.write( "start by multiplying two pairs of those binomials\r\n" );
document.write( "by FOIL.  We have a choice of \r\n" );
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document.write( "1. multiplying the 1st by the 2nd and the 3rd by \r\n" );
document.write( "the 4th, or\r\n" );
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document.write( "2. multiplying the 1st by the 3rd and the 2nd by\r\n" );
document.write( "the 4th, or\r\n" );
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document.write( "3. multiplying the 1st by the 4th and the 2nd by\r\n" );
document.write( "the 3rd, or\r\n" );
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document.write( " We investigate the 3 possibilities.\r\n" );
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document.write( "If we multiply the 1st by the 2nd and the 3rd by \r\n" );
document.write( "the 4th the middle terms would be -3x and 7x.  \r\n" );
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document.write( "If we multiply the 1st by the 3rd and the 2nd by \r\n" );
document.write( "the 4th, the middle terms would be 2x and 2x.\r\n" );
document.write( "Both the same.\r\n" );
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document.write( "If we multiply the 1st by the 4th and the 2nd by \r\n" );
document.write( "the 3rd the middle terms would be 3x and x.\r\n" );
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document.write( "The 2nd way is the best plan for we get the \r\n" );
document.write( "same middle term. \r\n" );
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document.write( "Rearrange the factors so that the ones we multiply\r\n" );
document.write( "will be side by side:\r\n" );
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document.write( "(x-1)(x+3)(x-2)(x+4)+4\r\n" );
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document.write( "(x²+2x-3)(x²+2x-8)+4\r\n" );
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document.write( "Let u = x²+2x\r\n" );
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document.write( "(u-3)(u-8)+4\r\n" );
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document.write( "u²-11u+24+4\r\n" );
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document.write( "u²-11u+28\r\n" );
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document.write( "(u-7)(u-4)\r\n" );
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document.write( "Replace u by x²+2x\r\n" );
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document.write( "(x²+2x-7)(x²+2x-4)\r\n" );
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document.write( "Edwin
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