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2 answers

I don't believe it can. It should only have four "turning points." It can have at most five "zeros," where the graph meets the x-axis, because we could write a five-degree polynomial:
(x-a)(x-b)(x-c)(x-d)(x-e)
which has zeros a, b, c, d, and e.
But the graph then can only "turn around" four times to hit each of these zeros.

2007-03-08 12:55:49 · answer #1 · answered by Ben 6 · 0 0

Take this 5th degree polynomial as an example. (x-1)(x-2)(x-3)(x-4)(x-5)=0. Its solutions are x=1,2,3,4 and 5. So it must have at least 4 turning points to cross the x axis this many times.

2007-03-08 12:52:11 · answer #2 · answered by zee_prime 6 · 0 0

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