document.write( "Question 25395: Factor:12w^2+10w-8\r
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Algebra.Com's Answer #13630 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
Factor:12w^2+10w-8 \r\n" );
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document.write( "Here's one just like it.  Use it as a model to\r\n" );
document.write( "factor yours by:\r\n" );
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document.write( "40x²+50x-15\r\n" );
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document.write( "First factor out the common factor of 5\r\n" );
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document.write( "5(8x²+10x-3)\r\n" );
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document.write( "Now to factor what is in the parentheses\r\n" );
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document.write( "Multiply the absolute values of the coefficient of the first \r\n" );
document.write( "term and last term\r\n" );
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document.write( "8·3 = 24\r\n" );
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document.write( "Think of two positive integers that have product 24 and\r\n" );
document.write( "which have DIFFERENCE of the absolute value of the coefficient \r\n" );
document.write( "of the middle term, 10.  [Note: it's DIFFERENCE when the last\r\n" );
document.write( "sign is -, but it's SUM when it's +] \r\n" );
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document.write( "These as 12 and 2.\r\n" );
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document.write( "Now use 12 and 2 to rewrite the middle term + 10x.\r\n" );
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document.write( "+10x = +12x - 2x\r\n" );
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document.write( "So replace +10x by +12x - 2x\r\n" );
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document.write( "5[8x²+12x-2x-3]\r\n" );
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document.write( "Now factor by grouping.  \r\n" );
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document.write( "Factor the first two terms by taking out 4x\r\n" );
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document.write( "5[4x(2x+3)-2x-3]\r\n" );
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document.write( "Now factor the last two terms.  You can take out a -1.\r\n" );
document.write( "[Always factor out a negative number if the third term is\r\n" );
document.write( "negative\r\n" );
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document.write( "5[4x(2x+3)-1(2x+3)]\r\n" );
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document.write( "Now factor out (2x+3)\r\n" );
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document.write( "5[(2x+3)(4x-1)]\r\n" );
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document.write( "Now dispense with the brackets.\r\n" );
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document.write( "5(2x+3)(4x-1)\r\n" );
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document.write( "The answer to your problem is either\r\n" );
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document.write( "2(3x+4)(2x-1) or 2(2x-1)(3x+4)\r\n" );
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document.write( "Edwin\r\n" );
document.write( "AnlytcPhil@aol.com
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