To determine the possible types and numbers of roots for the polynomial F(x)=−2x5+3x3+2x2−x−3 F(x) = -2x^5 + 3x^3 + 2x^2 - x - 3 F(x)=−2x5+3x3+2x2−x−3, we can use Descartes' Rule of Signs. This rule helps us find the number of positive and negative real roots by analyzing the sign changes in the polynomial and its transformation. Additionally, the Fundamental Theorem of Algebra tells us that the total number of roots (including complex roots) is equal to the degree of the polynomial, which is 5 in this case.
Using Descartes' Rule of Signs, we analyze the polynomial F(x)=−2x5+3x3+2x2−x−3 F(x) = -2x^5 + 3x^3 + 2x^2 - x - 3 F(x)=−2x5+3x3+2x2−x−3. The number of sign changes in F(x) F(x) F(x) is 2, indicating that there are either 2 or 0 positive real roots.
Next, we evaluate F(−x)=2x5−3x3+2x2+x−3 F(-x) = 2x^5 - 3x^3 + 2x^2 + x - 3 F(−x)=2x5−3x3+2x2+x−3. The number of sign changes in F(−x) F(-x) F(−x) is 3, which suggests there are either 3, 1, or no negative real roots.
The degree of the polynomial F(x) F(x) F(x) is 5. The total number of roots (including complex roots) must equal the degree. Given the findings from the previous steps, we have:
Thus, the number of imaginary roots can be calculated as: Imaginary Roots=Degree−(Positive Roots+Negative Roots)=5−(2+3)=0 \text{Imaginary Roots} = \text{Degree} - (\text{Positive Roots} + \text{Negative Roots}) = 5 - (2 + 3) = 0 Imaginary Roots=Degree−(Positive Roots+Negative Roots)=5−(2+3)=0
The possible types and numbers of roots for the function F(x) F(x) F(x) are:
Thus, the answer is: Positive: 2, Negative: 3, Imaginary: 0 \boxed{\text{Positive: } 2, \text{ Negative: } 3, \text{ Imaginary: } 0} Positive: 2, Negative: 3, Imaginary: 0
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