Algebra.Com's Answer #84935 by solver91311(24713)![\"\"](\"/images/stars/stars-13.gif\")  
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Your analysis has a small error.\r
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document.write( "if  , then  , which simplifies to the original function  because  for all even n.\r
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document.write( "Therefore,  has 1 sign change, so there is a maximum of 1 positive real root, and  also has 1 sign change, so there is a maximum of 1 negtative real root.\r
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document.write( "This just means that you have to do some more work because the analysis doesn't say anything about whether you definitely have any real roots, nor, if you do, whether they are rational or not.\r
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document.write( "Your rational root test analysis was correct. The only possible factors of 2 are 1 and 2, and the lead coefficient is 1, so your possible rational factors are -2, -1, 1, 2.\r
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document.write( "You could just test all four of those possibilities, but I prefer to look at it this way:\r
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document.write( "Let  . Now if , you can say that  \r
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document.write( "This factors rather nicely to  , but now we can substitute  back in for  and get:\r
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document.write( "We know that  if and only if  or  . Since we are looking for rational zeros, we can eliminate because this factor has complex roots with irrational coefficients,  and  .\r
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document.write( "But factors to  which gives us two real and rational roots, namely  and  .\r
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document.write( "Let's look at a graph of the original function and see if the result makes sense.\r
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document.write( "Yep. This is a 4th degree polynomial function, so we know that it has to have 4 roots, but the graph clearly intersects the x-axis in only two places, (-1,0) and (1,0), as we expected. That tells us that there are only 2 real roots and the other two have to be a conjugate pair of complex numbers.\r
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document.write( "Hope that helps,
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document.write( "John \n" );
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