document.write( "Question 127400: I have to simplify the polynomial and i am really confused. This problem is so different then the examples given in the text...\r
\n" ); document.write( "\n" ); document.write( "\"3x%28x%5E5%29%5E2\" divided by \"6x+%5E3%28x%5E2%29%5E4+\"\r
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Algebra.Com's Answer #93443 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
Given:
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\n" ); document.write( "\"3x%28x%5E5%29%5E2\" divided by \"6x+%5E3%28x%5E2%29%5E4+\"
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\n" ); document.write( "or
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\n" ); document.write( "\"%283x%28x%5E5%29%5E2%29%2F%286x+%5E3%28x%5E2%29%5E4+%29\"
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\n" ); document.write( "When you square \"x%5E5\" you can think of this in two ways. You multiply the exponent 2
\n" ); document.write( "times the exponent 5 to get \"x%5E10\" or you multiply \"x%5E5\" times itself to get:
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\n" ); document.write( "\"x%5E5+%2Ax%5E5+=+x%5E%285%2B5%29+=+x%5E10\"
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\n" ); document.write( "In either case, you substitute \"x%5E10\" for \"%28x%5E5%29%5E2\" in the numerator and the problem
\n" ); document.write( "then becomes:
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\n" ); document.write( "\"%283x%28x%5E10%29%29%2F%286x+%5E3%28x%5E2%29%5E4%29\"
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\n" ); document.write( "Now let's work on the term \"%28x%5E2%29%5E4\" in the denominator. If you multiply the two exponents
\n" ); document.write( "you get \"%28x%5E2%29%5E4+=+x%5E8\" which is also what you get if you multiply \"x%5E2\" by itself
\n" ); document.write( "4 times. Substituting this into the denominator makes the problem become:
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\n" ); document.write( "\"%283x%28x%5E10%29%29%2F%286x+%5E3%28x%5E8%29%29\"
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\n" ); document.write( "Back to the numerator. If you multiply \"3x\" times \"x%5E10\" you add the exponents of
\n" ); document.write( "the x terms and you get \"3x%5E11\" for the numerator.
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\n" ); document.write( "Then to the denominator. In multiplying \"x%5E3\" by \"x%5E8\" you get \"x%5E11\" by
\n" ); document.write( "adding the exponents. Then with the factor 6 the denominator becomes \"6x%5E11\". With
\n" ); document.write( "these reductions of the numerator and the denominator the problem then is reduced to:
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\n" ); document.write( "\"%283x%5E11%29%2F%286x%5E11%29\"
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\n" ); document.write( "Dividing the \"x%5E11\" of the numerator by the \"x%5E11\" of the denominator results in
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\n" ); document.write( "\"x%5E11%2Fx%5E11+=+x%5E%2811-11%29+=+x%5E0+=+1\"
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\n" ); document.write( "[or you could just view this as canceling the \"x%5E11\" of the numerator with the counterpart
\n" ); document.write( "\"x%5E11\" in the denominator.] This reduction leaves you with:
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\n" ); document.write( "\"3%2F6+=+1%2F2\"
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\n" ); document.write( "And so, the answer to this problem is that it all simplifies down to \"1%2F2\"
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\n" ); document.write( "Hope this helps you to understand the problem and one way you might solve it.
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