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Use Descartes' Rule of Signs to determine all possible combinations of real and imaginary zeros for the polynomial [tex]f(x)=-3x^4+4x^2+x-1[/tex].

Select all possible combinations of zeros:

A. 2 positive real zeros, 2 negative real zeros, 0 imaginary zeros
B. 2 positive real zeros, 0 negative real zeros, 2 imaginary zeros
C. 0 positive real zeros, 2 negative real zeros, 2 imaginary zeros
D. 0 positive real zeros, 0 negative real zeros, 4 imaginary zeros
E. 2 positive real zeros, 0 negative real zeros, 4 imaginary zeros
F. 3 positive real zeros, 1 negative real zeros, 0 imaginary zeros
G. 1 positive real zeros, 3 negative real zeros, 0 imaginary zeros
H. 4 positive real zeros, 0 negative real zeros, 0 imaginary zeros
I. 0 positive real zeros, 4 negative real zeros, 0 imaginary zeros

Sagot :

GinnyAnsweridn
To determine the possible combinations of real and imaginary zeros for the polynomial [tex]\( f(x) = -3x^4 + 4x^2 + x - 1 \)[/tex] using Descartes' Rule of Signs, we need to follow these steps:

1. Identify sign changes in [tex]\( f(x) \)[/tex]:

The polynomial [tex]\( f(x) \)[/tex] has the coefficients [tex]\([-3, 0, 4, 1, -1]\)[/tex].

- From [tex]\(-3\)[/tex] to [tex]\(0\)[/tex]: No sign change.
- From [tex]\(0\)[/tex] to [tex]\(4\)[/tex]: No sign change.
- From [tex]\(4\)[/tex] to [tex]\(1\)[/tex]: No sign change.
- From [tex]\(1\)[/tex] to [tex]\(-1\)[/tex]: Sign change.

Here, we have one sign change in [tex]\( f(x) \)[/tex], which indicates that there is one possible positive real root.

2. Identify sign changes in [tex]\( f(-x) \)[/tex]:

For [tex]\( f(-x) \)[/tex], we change the signs of the coefficients of odd powers of [tex]\( x \)[/tex]:

[tex]\[ f(-x) = -3(-x)^4 + 4(-x)^2 + (-x)- 1 = -3x^4 + 4x^2 - x - 1 \][/tex]

The polynomial [tex]\( f(-x) \)[/tex] has the coefficients [tex]\([-3, 0, 4, -1, -1]\)[/tex].

- From [tex]\(-3\)[/tex] to [tex]\(0\)[/tex]: No sign change.
- From [tex]\(0\)[/tex] to [tex]\(4\)[/tex]: No sign change.
- From [tex]\(4\)[/tex] to [tex]\(-1\)[/tex]: Sign change.
- From [tex]\(-1\)[/tex] to [tex]\(-1\)[/tex]: No sign change.

Here, we have one sign change in [tex]\( f(-x) \)[/tex], which indicates that there is one possible negative real root.

3. Possible combinations of zeros:

Given the degree of [tex]\( f(x) \)[/tex] is 4 (quartic polynomial), the total number of zeros (real and imaginary) must equal 4. We combine the possibilities for positive real zeros, negative real zeros, and the remaining imaginary zeros:

- 1 positive real zero and 1 negative real zero:
Remaining zeros [tex]\( = 4 - 1 - 1 = 2 \)[/tex] imaginary zeros.
Combination: 1 positive real zero, 1 negative real zero, 2 imaginary zeros.

However, after reviewing all options from the provided list, none of them matches our findings from Descartes' rule of signs entirely.

Thus, the final conclusion indicates:
None of the given combinations provide a valid scenario for the polynomial based on the Descartes' rule of signs analysis.
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