According to Descartes' rule of signs, there can be either two real positive zeros (roots), or none.
According to the same rule, by consideration of sign changes in f(-x), there can be either two real negative zeros, or none.
Overall, then, there could be 0, 2, 4, 6 or even 8 imaginary zeros (occurring as conjugate pairs), depending on however many real roots there turn out to be; for the latter, I recommend exploring the function on a graphing calculator.
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Determine the possible number of Positive and Negative Real Zeros of the polynomial using Descartes' Rule of Signs
Possible Answers:
Positive zeroes:
Negative zeroes:
Positive zeroes:
Negative zeroes:
Positive zeroes:
Negative zeroes:
Positive zeroes:
Negative zeroes:
Correct answer:
Positive zeroes:
Negative zeroes:
Explanation:
In order to determine the positive number of real zeroes, we must count the number of sign changes in the coefficients of the terms of the polynomial. The number of real zeroes can then be any positive difference of that number and a positive multiple of two.
For the function
there are four sign changes. As such, the number of positive real zeroes can be
In order to determine the positive number of real zeroes, we must count the number of sign changes in the coefficients of the terms of the polynomial after substituting for The number of real zeroes can then be any positive difference of that number and a positive multiple of two.
After substituting we get
and there is one sign change.
As such, there can only be one negative root.
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