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grei
Posts:
37
Registered:
11/27/12
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Re: How can I know how many real roots this polynomial has?
Posted:
Jan 16, 2013 10:27 AM
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The "intermediate value theorem" says that if P(a)< 0 and P(B)> 0 then there exist [b]at least[/b] one x between a and b such that P(x)= 0. But it will not tell you how many!
Similarly, we can use DeCarte's rule: If, in counting positive and negative signs we find that there are "n sign changes" as we go from highest degree to lowest, then there are no more than n positive real roots and the actual nummber must differ from n by a multiple of two.
Here, assuming that "?" between x^7 and x^5 was a supposed to be a "+", there are NO changes of sign and so no positive real roots (obviously- all values are positive and cannot add to 0). If we change the sign on x, swapping positive and negative values, we change the sign on odd powers, getting -x^7- 10x^5- 15x+ 5= 0 so there is exactly one sign change and, since there is no non-negative number less than that by a multiple of two, there must be exactly one negative root. That is, DesCartes' rule of signs tells us this polynomial, x^7+ 10x^5+ 15x+ 5, has exactly one real (negative) zero.
But if that "?" is a negative, the number of changes in sign, + - + +, is 2 so there could be either 2 or 0 positive roots. Swapping positive and negative x gives the polynomial -x^7+ 10x^5- 15x +5 which has 3 changes in sign and so 3 or 1 negative root. So for x^7- 10x^5+ 15x+ 5, all we can say is that there may be 1, 3, or 5 real roots.
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