Question
State the possible number of positive real zeroes, negative real zeroes, and imaginary zeroes of each function.
p(x)=2x^4-2x^3+2x^2-x-1
p(x)=2x^4-2x^3+2x^2-x-1
Answers
GPT-4o mini
To determine the possible number of positive real zeros, negative real zeros, and imaginary zeros of the polynomial \( p(x) = 2x^4 - 2x^3 + 2x^2 - x - 1 \), we can use Descartes' Rule of Signs.
### Step 1: Positive Real Zeros
We first look at \( p(x) \) to count the sign changes:
- \( p(x) = 2x^4 \) (positive)
- \( -2x^3 \) (negative)
- \( 2x^2 \) (positive)
- \( -x \) (negative)
- \( -1 \) (negative)
The sequence of signs is: \( +, -, +, -, - \).
Counting the sign changes:
1. From \( + \) to \( - \) (1 change)
2. From \( - \) to \( + \) (2nd change)
3. From \( + \) to \( - \) (3rd change)
Total sign changes = 3.
By Descartes’ Rule of Signs, the number of positive real zeros is equal to the number of sign changes, or that number decreased by an even number. Thus, the possible number of positive real zeros is:
- 3, 1 (3 - 2).
### Step 2: Negative Real Zeros
Next, to find negative real zeros, we evaluate \( p(-x) \).
\[
p(-x) = 2(-x)^4 - 2(-x)^3 + 2(-x)^2 - (-x) - 1 = 2x^4 + 2x^3 + 2x^2 + x - 1
\]
Now, we examine the sign changes in \( p(-x) \):
- \( 2x^4 \) (positive)
- \( +2x^3 \) (positive)
- \( +2x^2 \) (positive)
- \( +x \) (positive)
- \( -1 \) (negative)
The sequence of signs is: \( +, +, +, +, - \).
Counting the sign changes:
- There is 1 sign change (from positive to negative).
Thus, by Descartes' Rule of Signs, the possible number of negative real zeros is 1.
### Step 3: Imaginary Zeros
A polynomial of degree \( n \) has exactly \( n \) zeros in the complex number system (including multiplicities), according to the Fundamental Theorem of Algebra. For our polynomial \( p(x) \), which is degree 4:
\[
\text{Total zeros} = 4.
\]
Now, to summarize:
- Possible positive real zeros: 3 or 1
- Possible negative real zeros: 1
- Possible imaginary zeros: Since the total number of zeros is 4, and we have identified a maximum of 3 positive and 1 negative real zero, the imaginary zeros must account for the remaining zeros.
### Conclusion:
The possible number of zeros for the polynomial \( p(x) \) is:
- **Positive real zeros**: 3 or 1
- **Negative real zeros**: 1
- **Imaginary zeros**: 0 or 2 (depending on the number of positive zeros).
This gives a complete overview based on Descartes' Rule of Signs.
### Step 1: Positive Real Zeros
We first look at \( p(x) \) to count the sign changes:
- \( p(x) = 2x^4 \) (positive)
- \( -2x^3 \) (negative)
- \( 2x^2 \) (positive)
- \( -x \) (negative)
- \( -1 \) (negative)
The sequence of signs is: \( +, -, +, -, - \).
Counting the sign changes:
1. From \( + \) to \( - \) (1 change)
2. From \( - \) to \( + \) (2nd change)
3. From \( + \) to \( - \) (3rd change)
Total sign changes = 3.
By Descartes’ Rule of Signs, the number of positive real zeros is equal to the number of sign changes, or that number decreased by an even number. Thus, the possible number of positive real zeros is:
- 3, 1 (3 - 2).
### Step 2: Negative Real Zeros
Next, to find negative real zeros, we evaluate \( p(-x) \).
\[
p(-x) = 2(-x)^4 - 2(-x)^3 + 2(-x)^2 - (-x) - 1 = 2x^4 + 2x^3 + 2x^2 + x - 1
\]
Now, we examine the sign changes in \( p(-x) \):
- \( 2x^4 \) (positive)
- \( +2x^3 \) (positive)
- \( +2x^2 \) (positive)
- \( +x \) (positive)
- \( -1 \) (negative)
The sequence of signs is: \( +, +, +, +, - \).
Counting the sign changes:
- There is 1 sign change (from positive to negative).
Thus, by Descartes' Rule of Signs, the possible number of negative real zeros is 1.
### Step 3: Imaginary Zeros
A polynomial of degree \( n \) has exactly \( n \) zeros in the complex number system (including multiplicities), according to the Fundamental Theorem of Algebra. For our polynomial \( p(x) \), which is degree 4:
\[
\text{Total zeros} = 4.
\]
Now, to summarize:
- Possible positive real zeros: 3 or 1
- Possible negative real zeros: 1
- Possible imaginary zeros: Since the total number of zeros is 4, and we have identified a maximum of 3 positive and 1 negative real zero, the imaginary zeros must account for the remaining zeros.
### Conclusion:
The possible number of zeros for the polynomial \( p(x) \) is:
- **Positive real zeros**: 3 or 1
- **Negative real zeros**: 1
- **Imaginary zeros**: 0 or 2 (depending on the number of positive zeros).
This gives a complete overview based on Descartes' Rule of Signs.