document.write( "Question 468180: I'm completely lost...
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Algebra.Com's Answer #321250 by Edwin McCravy(20059) $\"\"$ $\"About$
You can put this solution on YOUR website!
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document.write( "3x²+5x+2\r\n" );
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document.write( "Multiply the 3 by the 2 ignoring signs.  Get 6.\r\n" );
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document.write( "Write down all the ways to have two positive integers\r\n" );
document.write( "which have product 6, starting with 6*1\r\n" );
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document.write( "6*1\r\n" );
document.write( "3*2\r\n" );
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document.write( "Since the last sign in 3x²+5x+2 is +, ADD them,\r\n" );
document.write( "and place the SUM out beside that:\r\n" );
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document.write( "6*1    6+1=7\r\n" );
document.write( "3*2    3+2=5\r\n" );
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document.write( "Now, again ignoring signs, we find in that list of\r\n" );
document.write( "sums the coefficient of the middle term in 3x²+5x+2\r\n" );
document.write( "which is 5\r\n" );
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document.write( "So we replace the number 5 by 3+2\r\n" );
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document.write( "3x²+5x+2\r\n" );
document.write( "3x²+(3+2)x+2 \r\n" );
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document.write( "Then we distribute to remove the parentheses:\r\n" );
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document.write( "3x²+3x+2x+2\r\n" );
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document.write( "Factor the first two terms 3x²+3x by taking out the\r\n" );
document.write( "greatest common factor, getting 3x(x+1)\r\n" );
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document.write( "Factor the last two terms +2x+2 by taking out the\r\n" );
document.write( "greatest common factor, +2, getting +2(x+1)\r\n" );
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document.write( "So we have\r\n" );
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document.write( "3x(x+1)+2(x+1)\r\n" );
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document.write( "Notice that there is a common factor, (x+1)\r\n" );
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document.write( "3x(x+1)+2(x+1)\r\n" );
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document.write( "which we can factor out leaving the 3x and the +2 to put \r\n" );
document.write( "in parentheses:\r\n" );
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document.write( "(x+1)(3x+2)\r\n" );
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document.write( "------------------------------\r\n" );
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document.write( "5x²-14x-3\r\n" );
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document.write( "Multiply the 5 by the 3 ignoring signs.  Get 15\r\n" );
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document.write( "Write down all the ways to have two positive integers\r\n" );
document.write( "which have product 15, starting with 15*1\r\n" );
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document.write( "15*1\r\n" );
document.write( " 5*3\r\n" );
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document.write( "Since the last sign in 5x²-14x-3 is -, SUBTRACT them,\r\n" );
document.write( "and place the DIFFERENCE out beside that:\r\n" );
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document.write( "15*1   15-1=14\r\n" );
document.write( " 5*3    5-3=2\r\n" );
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document.write( "Now, again ignoring signs, we find in that list of\r\n" );
document.write( "sums the coefficient of the middle term in 5x²-14x-3\r\n" );
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document.write( "So we replace the number 14 by 15-1\r\n" );
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document.write( "5x²-14x-3\r\n" );
document.write( "5x²-(15-1)x-3\r\n" );
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document.write( "Then we distribute to remove the parentheses:\r\n" );
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document.write( "5x²-15x+1x-3\r\n" );
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document.write( "Factor the first two terms 5x²-15x by taking out the\r\n" );
document.write( "greatest common factor, 5x, getting 5x(x-3)\r\n" );
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document.write( "Factor the last two terms +1x-3 by taking out the\r\n" );
document.write( "greatest common factor, +1, getting +1(x-3)\r\n" );
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document.write( "So we have\r\n" );
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document.write( "5x(x-3)+1(x-3)\r\n" );
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document.write( "Notice that there is a common factor, (x-3)\r\n" );
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document.write( "5x(x-3)+1(x-3)\r\n" );
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document.write( "which we can factor out leaving the 5x and the +1 to put \r\n" );
document.write( "in parentheses:\r\n" );
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document.write( "(x-3)(5x+1)\r\n" );
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document.write( "---------------------------\r\n" );
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document.write( "The others are done just like the first one.\r\n" );
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document.write( "I wrote a lesson on this method of factoring.  Go here:\r\n" );
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document.write( "http://www.algebra.com/algebra/homework/Expressions-with-variables/change-this-name32371.lesson\r\n" );
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document.write( "Edwin
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