document.write( "Question 102442This question is from textbook Intermediate Algebra
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Algebra.Com's Answer #74469 by Edwin McCravy(20064) $\"\"$ $\"About$
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document.write( " 6xy-4x-9y+6 \r\n" );
document.write( "  6y²-13y+6\r\n" );
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document.write( "First we must factor the numerator:\r\n" );
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document.write( "6xy - 4x - 9y + 6\r\n" );
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document.write( "Out of the first two terms we factor out 2x:\r\n" );
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document.write( "2x(3y - 2) - 9y + 6\r\n" );
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document.write( "Out of the last two terms we factor out -3\r\n" );
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document.write( "2x(3y - 2) - 3(3y - 2)\r\n" );
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document.write( "Now we notice that there is a common factor\r\n" );
document.write( "of (3y - 2).  So we factor that red factor\r\n" );
document.write( "out putting the black parts in parentheses:\r\n" );
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document.write( "(3y - 2)(2x - 3)\r\n" );
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document.write( "or writing it all in black:\r\n" );
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document.write( "(3y - 2)(2x - 3)\r\n" );
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document.write( "Next we must factor the denominator:\r\n" );
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document.write( "6y² - 13y + 6\r\n" );
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document.write( "Multiply the red 6 by the purple 6, getting 36\r\n" );
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document.write( "Think of two integers which have product 36 and SUM the green 13.\r\n" );
document.write( "(Note: the reason it's SUM and not DIFFERENCE is because the last\r\n" );
document.write( "sign in the trinomial is +. If it had been - we would have said\r\n" );
document.write( "\"DIFFERENCE\" here).\r\n" );
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document.write( "Anyway, two integers whose product is 36 and whose sum is 13 are\r\n" );
document.write( "9 and 4.  So we rewrite 13 as either 4 + 9 or 9 + 4, whichever you\r\n" );
document.write( "choose. I will choose 9 + 4:\r\n" );
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document.write( "So the denominator \r\n" );
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document.write( "    6y² - 13y + 6\r\n" );
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document.write( "becomes\r\n" );
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document.write( "    6y² - (9 + 4)y + 6\r\n" );
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document.write( "Distribute to remove the parentheses:\r\n" );
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document.write( "    6y² - 9y - 4y + 6\r\n" );
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document.write( "Factor 3y out of the first two terms:\r\n" );
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document.write( "   3y(2y - 3) - 4y + 6\r\n" );
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document.write( "Factor -2 out of the last two terms:\r\n" );
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document.write( "   3y(2y - 3) - 2(2y - 3)\r\n" );
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document.write( "Now we notice that there is a common factor\r\n" );
document.write( "of (2y - 3).  So we factor that red factor\r\n" );
document.write( "out putting the black parts in parentheses:\r\n" );
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document.write( "(2y - 3)(3y - 2)\r\n" );
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document.write( "or writing it all in black:\r\n" );
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document.write( "(2y - 3)(3y - 2)\r\n" );
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document.write( "Now putting the factored numerator over the factored denominator:\r\n" );
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document.write( " (3y - 2)(2x - 3) \r\n" );
document.write( " (2y - 3)(3y - 2)\r\n" );
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document.write( "Now we cancel the (3y - 2)'s\r\n" );
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document.write( "     1\r\n" );
document.write( " (3y - 2)(2x - 3) \r\n" );
document.write( " (2y - 3)(3y - 2)\r\n" );
document.write( "             1\r\n" );
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document.write( "and all that's left is \r\n" );
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document.write( " (2x - 3)\r\n" );
document.write( " (2y - 3)\r\n" );
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document.write( "We do not need the parentheses:\r\n" );
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document.write( " 2x - 3 \r\n" );
document.write( " 2y - 3  \r\n" );
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document.write( "Caution: do not try to cancel anything else\r\n" );
document.write( "because neither the 2's not the -3's are\r\n" );
document.write( "factors of the numerator and denominator. \r\n" );
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document.write( "Edwin
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