et P(x) = 2x4 ? 7x3 + x2 ? 18x + 3. Use Descartes\' Rule of Signs to determine h
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P(x) = 2x4 ? 7x3 + x2 ? 18x + 3. Use Descartes' Rule of Signs to determine how many positive and how many negative real zeros P can have. P(x) must have 0 positive real zeroes and can have 3 or 1 negative real zeroes. P(x) can have 4, 2, or 0 positive real zeroes and must have 0 negative real zeroes. P(x) can have 4, 3, 2, 1, or 0 positive real zeroes and must have 0 negative real zeroes. P(x) can have 4, 2, or 0 positive real zeroes and can have 3 or 1 negative real zeroes. P(x) must have 0 positive real zeroes and can have 4, 3, 2, 1, or 0 negative real zeroes.Explanation / Answer
Descartes' Rule of Signs will not tell you where the polynomial's zeroes are (you'll need to use the Rational Roots Test and synthetic division, or draw a graph, to actually find the roots), but the Rule will tell you how many roots you can expect.
First, I look at the polynomial as it stands, not changing the sign on x, so this is the "positive" case:
I draw little lines underneath to highlight where the signs change from positive to negative or from negative to positive from one term to the next
P(x) = 2x4 - 7x3 + x2 - 18x + 3. Then I count the number of changes.
There are four sign changes in the "positive" case. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial
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 two's to find the complete list of the possible number of zeroes. That is, while there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all).
Now I look at f (
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