document.write( "Question 1041277: Can u help me factorise this
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Algebra.Com's Answer #656246 by AnlytcPhil(1806)\"\" \"About 
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I'm thinking the x on the first term may have been a typo\r\n" );
document.write( "and you meant this:\r\n" );
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document.write( "23+2y + 24+y∙3 - 2y - 6\r\n" );
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document.write( "If so then out of the first two terms,\r\n" );
document.write( "factorise out the smaller power of 2 factor, \r\n" );
document.write( "which is 24+y by subtracting the\r\n" );
document.write( "exponents:  24+y(2(3+2y)-(4+y)+3) = 24+y(23+2y-4-y+3) = 24+y(2y-1+3) \r\n" );
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document.write( "24+y(2y-1+3) - 2y - 6\r\n" );
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document.write( "Out of the last two terms factorize out -2 which is the same\r\n" );
document.write( "as 21.  We do this also by subtracting exponents:\r\n" );
document.write( "-2y-6 = -21(2y-1+3) = -2(2y-1+3).\r\n" );
document.write( "Now we have:\r\n" );
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document.write( "24+y(2y-1+3) - 2(2y-1+3).\r\n" );
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document.write( "Now there is a common factor of (2y-1+3).\r\n" );
document.write( "Factorizing it out on the right, we have:\r\n" );
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document.write( "(24+y - 2)(2y-1+3)\r\n" );
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document.write( "Finally we can factorize out 2 from the first parentheses\r\n" );
document.write( "by subtracting exponents: 21(24+y-1 - 1) = 21(23+y - 1) = 2(23+y - 1)  \r\n" );
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document.write( "2(23+y - 1)(2y-1+3)\r\n" );
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document.write( "Edwin
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