http://i453.photobucket.com/albums/q...8/mathrev1.jpg

Not sure on this, so p1(x)/q(x) = x(x-1) = x^2 - x

p2(x)/q(x) = x^2 - 2x + 1

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- May 13th 2009, 06:16 AMmath_helpMinimum polynomial
http://i453.photobucket.com/albums/q...8/mathrev1.jpg

Not sure on this, so p1(x)/q(x) = x(x-1) = x^2 - x

p2(x)/q(x) = x^2 - 2x + 1 - May 13th 2009, 11:27 AMGamma?
The picture and text do not match up.

What is your question? - May 13th 2009, 01:03 PMmath_help
Sorry, wrong one :)