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There are three changes of sign in f(x). So there must be either: three positive roots, or, one positive root and one pair of complex conjugates. f(-x) = - 7x5 + 15x4 + x3 + 4x2 + 6x – 11 There are two changes of sign in f(-x). So there must be either two negative roots, or one pair of complex conjugates. (In fact there are two real negative roots, one positive real root and a pair of complex conjugates, but Descartes can't be that specific.) The rational roots theorem tells you to check for possible roots at ± (1, 3, 11, 33, 1/3, 11/3, 1/9, and 11/9). Descartes tells you that there are no negative roots, so you need to check only the positive values in the list. But in fact that f(x) has no rational (or real) zeros.
However, some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex and thus not graphable as x-intercepts. Because of this possibility, I have to count down by 3'sto find the complete list of the possible number of zeroes. That is, while there may be as many as 3 real zeroes, and there might also be zero (none at all). Now I look at f (–x) (that is, having changed the sign on x, so this is the "negative" case):