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grei
Posts:
116
Registered:
11/27/12


Re: How can I know how many real roots this polynomial has?
Posted:
Jan 16, 2013 10:27 AM


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 nonnegative 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.



