Question
Use Descartes’s Rule of Signs to describe the real zeroes of the function f left parenthesis x right parenthesis equals 2 x superscript 5 baseline minus x superscript 4 baseline minus 2 x cubed plus 4 x squared plus x minus 2.
(1 point)
Responses
The function has two or zero positive real zeroes and either three or one negative real zeroes.
The function has two or zero positive real zeroes and either three or one negative real zeroes.
The function has two or zero negative real zeroes and either three or one positive real zeroes.
The function has two or zero negative real zeroes and either three or one positive real zeroes.
The function has one positive real zeroes and either four or two negative real zeroes.
The function has one positive real zeroes and either four or two negative real zeroes.
The function has one negative real zeroes and either four or two positive real zeroes.
The function has one negative real zeroes and either four or two positive real zeroes.
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(1 point)
Responses
The function has two or zero positive real zeroes and either three or one negative real zeroes.
The function has two or zero positive real zeroes and either three or one negative real zeroes.
The function has two or zero negative real zeroes and either three or one positive real zeroes.
The function has two or zero negative real zeroes and either three or one positive real zeroes.
The function has one positive real zeroes and either four or two negative real zeroes.
The function has one positive real zeroes and either four or two negative real zeroes.
The function has one negative real zeroes and either four or two positive real zeroes.
The function has one negative real zeroes and either four or two positive real zeroes.
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Answers
GPT-4o mini
To analyze the real zeroes of the function \( f(x) = 2x^5 - x^4 - 2x^3 + 4x^2 + x - 2 \) using Descartes's Rule of Signs, we follow these steps:
1. **Positive Real Zeroes**: Evaluate \( f(x) \) and count the number of sign changes in \( f(x) \):
- Coefficients: \( 2, -1, -2, 4, 1, -2 \)
- Signs: \( +, -, -, +, +, - \)
- Sign changes:
- \( + \rightarrow - \) (from \( 2 \) to \( -1 \)) - 1 change
- \( - \rightarrow + \) (from \( -1 \) to \( 4 \)) - 2 changes
- \( + \rightarrow - \) (from \( 4 \) to \( -2 \)) - 3 changes
- Total sign changes = 3
According to Descartes’s Rule of Signs, the number of positive real zeroes is either equal to the number of sign changes or less than it by an even integer. Therefore, there could be 3, 1, or 0 positive real zeroes.
2. **Negative Real Zeroes**: Evaluate \( f(-x) \):
\[
f(-x) = 2(-x)^5 - (-x)^4 - 2(-x)^3 + 4(-x)^2 + (-x) - 2 = -2x^5 - x^4 + 2x^3 + 4x^2 - x - 2
\]
- Coefficients: \( -2, -1, 2, 4, -1, -2 \)
- Signs: \( -, -, +, +, -, - \)
- Sign changes:
- \( - \rightarrow + \) (from \( -2 \) to \( 2 \)) - 1 change
- \( + \rightarrow - \) (from \( 4 \) to \( -1 \)) - 2 changes
- Total sign changes = 2
Therefore, the number of negative real zeroes is either 2 or 0.
### Conclusion
From the analysis using Descartes's Rule of Signs:
- There are either 3, 1, or 0 positive real zeroes.
- There are either 2 or 0 negative real zeroes.
So the correct response would be:
**The function has three or one positive real zeroes and either two or zero negative real zeroes.**
However, given the options provided, the most fitting response is:
**The function has two or zero positive real zeroes and either three or one negative real zeroes.**
1. **Positive Real Zeroes**: Evaluate \( f(x) \) and count the number of sign changes in \( f(x) \):
- Coefficients: \( 2, -1, -2, 4, 1, -2 \)
- Signs: \( +, -, -, +, +, - \)
- Sign changes:
- \( + \rightarrow - \) (from \( 2 \) to \( -1 \)) - 1 change
- \( - \rightarrow + \) (from \( -1 \) to \( 4 \)) - 2 changes
- \( + \rightarrow - \) (from \( 4 \) to \( -2 \)) - 3 changes
- Total sign changes = 3
According to Descartes’s Rule of Signs, the number of positive real zeroes is either equal to the number of sign changes or less than it by an even integer. Therefore, there could be 3, 1, or 0 positive real zeroes.
2. **Negative Real Zeroes**: Evaluate \( f(-x) \):
\[
f(-x) = 2(-x)^5 - (-x)^4 - 2(-x)^3 + 4(-x)^2 + (-x) - 2 = -2x^5 - x^4 + 2x^3 + 4x^2 - x - 2
\]
- Coefficients: \( -2, -1, 2, 4, -1, -2 \)
- Signs: \( -, -, +, +, -, - \)
- Sign changes:
- \( - \rightarrow + \) (from \( -2 \) to \( 2 \)) - 1 change
- \( + \rightarrow - \) (from \( 4 \) to \( -1 \)) - 2 changes
- Total sign changes = 2
Therefore, the number of negative real zeroes is either 2 or 0.
### Conclusion
From the analysis using Descartes's Rule of Signs:
- There are either 3, 1, or 0 positive real zeroes.
- There are either 2 or 0 negative real zeroes.
So the correct response would be:
**The function has three or one positive real zeroes and either two or zero negative real zeroes.**
However, given the options provided, the most fitting response is:
**The function has two or zero positive real zeroes and either three or one negative real zeroes.**