document.write( "Question 66380This question is from textbook Alegbra and Trigonometry
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Algebra.Com's Answer #47096 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "      x4 - 1\r\n" );
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document.write( " x3 + x2 + x + 1\r\n" );
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document.write( "Factor the numerator completely:\r\n" );
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document.write( "                x4 - 1\r\n" );
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document.write( "That's the difference of two perfect squares, so\r\n" );
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document.write( "             (x2 - 1)(x2 + 1)\r\n" );
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document.write( "The first factor is also the difference of two\r\n" );
document.write( "perfect squares, so the complete factorization is:\r\n" );
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document.write( "         (x - 1)(x + 1)(x2 + 1)\r\n" );
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document.write( "(Note that you CANNOT factor the SUM of two perfect\r\n" );
document.write( " squares, only the DIFFERENCE of two perfect squares)\r\n" );
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document.write( "Factor the denominator completely, by grouping:\r\n" );
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document.write( " x3 + x2 + x + 1\r\n" );
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document.write( "Factor x2 out of the first two terms:\r\n" );
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document.write( " x2(x + 1) + x + 1\r\n" );
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document.write( "Factor 1 out of the last two terms:\r\n" );
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document.write( " x2(x + 1) + 1(x + 1)\r\n" );
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document.write( "Now factor (x + 1) out of both terms:\r\n" );
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document.write( " (x + 1)(x2 + 1)\r\n" );
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document.write( "Now return to the original expression:\r\n" );
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document.write( "      x4 - 1\r\n" );
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document.write( " x3 + x2 + x + 1\r\n" );
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document.write( "and replace the numerator and denominator\r\n" );
document.write( "by their complete factorizations:\r\n" );
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document.write( "  (x - 1)(x + 1)(x2 + 1)  \r\n" );
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document.write( "    (x + 1)(x2 + 1)\r\n" );
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document.write( "Cancel the (x + 1)'s and the (x2 + 1)'s\r\n" );
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document.write( "            1       1\r\n" );
document.write( "  (x - 1)(x + 1)(x² + 1)  \r\n" );
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document.write( "    (x + 1)(x² + 1)\r\n" );
document.write( "       1       1\r\n" );
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document.write( "and all you have left is a measly:\r\n" );
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document.write( "      x - 1\r\n" );
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document.write( "Edwin

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