document.write( "Question 206554This question is from textbook Introductory Algebra
\n" ); document.write( ": Please help me factor this equation: \"x%5E%282n%2B1%29-2x%5E%28n%2B1%29%2Bx\"
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Algebra.Com's Answer #156095 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
The last solution is correct, but here's why \"x%5E%282n%29-2x%5En%2B1\" factors to \"%28x%5En-1%29%5E2\"...\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E%282n%29-2x%5En%2B1\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"%28x%5En%29%5E2-2x%5En%2B1\" Rewrite \"x%5E%282n%29\" as \"%28x%5En%29%5E2\". Note: \"x%5E%28y%2Az%29=%28x%5E%28y%29%29%5Ez\"\r
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\n" ); document.write( "\n" ); document.write( "Now let \"y=x%5En\" (a substitution to make things easier)\r
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\n" ); document.write( "\n" ); document.write( "\"y%5E2-2y%2B1\" Replace each \"x%5En\" with 'y'\r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression \"y%5E2-2y%2B1\", we can see that the first coefficient is \"1\", the second coefficient is \"-2\", and the last term is \"1\".\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient \"1\" by the last term \"1\" to get \"%281%29%281%29=1\".\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to \"1\" (the previous product) and add to the second coefficient \"-2\"?\r
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\n" ); document.write( "\n" ); document.write( "To find these two numbers, we need to list all of the factors of \"1\" (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of \"1\":\r
\n" ); document.write( "\n" ); document.write( "1\r
\n" ); document.write( "\n" ); document.write( "-1\r
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\n" ); document.write( "\n" ); document.write( "Note: list the negative of each factor. This will allow us to find all possible combinations.\r
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\n" ); document.write( "\n" ); document.write( "These factors pair up and multiply to \"1\".\r
\n" ); document.write( "\n" ); document.write( "1*1 = 1
\n" ); document.write( "(-1)*(-1) = 1\r
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\n" ); document.write( "\n" ); document.write( "Now let's add up each pair of factors to see if one pair adds to the middle coefficient \"-2\":\r
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First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2
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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers \"-1\" and \"-1\" add to \"-2\" (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers \"-1\" and \"-1\" both multiply to \"1\" and add to \"-2\"\r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term \"-2y\" with \"-y-y\". Remember, \"-1\" and \"-1\" add to \"-2\". So this shows us that \"-y-y=-2y\".\r
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\n" ); document.write( "\n" ); document.write( "\"y%5E2%2Bhighlight%28-y-y%29%2B1\" Replace the second term \"-2y\" with \"-y-y\".\r
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\n" ); document.write( "\n" ); document.write( "\"%28y%5E2-y%29%2B%28-y%2B1%29\" Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( "\"y%28y-1%29%2B%28-y%2B1%29\" Factor out the GCF \"y\" from the first group.\r
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\n" ); document.write( "\n" ); document.write( "\"y%28y-1%29-1%28y-1%29\" Factor out \"1\" from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.\r
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\n" ); document.write( "\n" ); document.write( "\"%28y-1%29%28y-1%29\" Combine like terms. Or factor out the common term \"y-1\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28y-1%29%5E2\" Condense the terms.\r
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\n" ); document.write( "\n" ); document.write( "\"%28x%5En-1%29%5E2\" Plug in \"y=x%5En\"\r
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\n" ); document.write( "\n" ); document.write( "So \"x%5E%282n%29-2x%5En%2B1\" factors to \"%28x%5En-1%29%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "In other words, \"x%5E%282n%29-2x%5En%2B1=%28x%5En-1%29%5E2\".\r
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