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Basically, you have 2 steps: 1. Look at the "sign changes" of the coefficients of f(x). So above, your function has coefficient signs in the following pattern: + + - - so there is ONE sign change. Now you "count down in pairs" - but since you result was 1, there is no counting down to do. You have AT MOST 1 POSITIVE ROOT. 2. Now look at the sign changes of the coefficients of f(-x)... just plug in (-x) and simplify the signs on the coefficients (remember: where the power is odd you will have a sign change; where the power is even you wont). I'll let you do this one on your own, but again, this gives you the max number of NEGATIVE roots; you may that that many, or that many - 2, or that many -4, etc.... you count down by 2's. Notice that the max of the positive roots and the max of the negative roots will always sum to the degree of the polynomial. So in the end, you can summarize the possible #'s of positive, negative, and complex roots.
Just as a "for example", suppose I have a 5th degree polynomial, and: For f(x) I have 3 sign changes (3 or 1 positive roots) For f(-x) I have 2 sign changes (2 or 0 negative roots) Then the possible combos of root types are: 3 positive, 2 negative, 0 complex roots 3 positive, 0 negative, 2 complex roots 1 positive, 2 negative, 2 complex roots 1 positive, 0 negative, 4 complex roots
thank you I had problems b/c when I was plugging in potential roots into a synthetic division chart I was getting confused