Answer :
Sure, let's break down the solution for each polynomial step-by-step:
### For the function [tex]\( f(x) = 4x^6 - 3x^4 - 16x^2 + 12 \)[/tex]:
1. Determine the possible number of imaginary zeros:
- The degree of the polynomial is 6. Imaginary (or complex) zeros appear in conjugate pairs. Thus, the polynomial can have 0, 2, 4, or 6 imaginary zeros.
2. Determine the possible number of positive zeros:
- Use Descartes' Rule of Signs on [tex]\( f(x) \)[/tex].
- Look at the signs of the coefficients: [tex]\( [4, -3, -16, 12] \)[/tex] corresponds to signs [tex]\( [+, -, -, +] \)[/tex].
- Count the sign changes in the sequence: [tex]\( + \to - \to - \to + \)[/tex]. There are 3 sign changes.
- The possible number of positive zeros is therefore 3 or 1 (subtract multiples of 2 from the number of sign changes).
3. Determine the possible number of negative zeros:
- Evaluate [tex]\( f(-x) \)[/tex] instead of [tex]\( f(x) \)[/tex].
- [tex]\( f(-x) = 4x^6 - 3x^4 + 16x^2 + 12 \)[/tex].
- The signs are [tex]\( [+, -, +, +] \)[/tex].
- Count the sign changes: [tex]\( + \to - \to + \to + \)[/tex]. There are 2 sign changes.
- The possible number of negative zeros is 2 or 0 (subtract multiples of 2).
### For the function [tex]\( f(x) = 25x^5 - 15x^4 + 145x^3 - 87x^2 + 100x - 60 \)[/tex]:
1. Determine the possible number of imaginary zeros:
- The degree of the polynomial is 5. Imaginary zeros can again appear in pairs, so we can have 0, 2, or 4 imaginary zeros. However, we cannot have 6 here since the degree is not even.
2. Determine the possible number of positive zeros:
- Apply Descartes' Rule of Signs to [tex]\( f(x) \)[/tex].
- The sign sequence of the coefficients is [tex]\( [25, -15, 145, -87, 100, -60] \)[/tex], resulting in signs [tex]\( [+, -, +, -, +, -] \)[/tex].
- Count the sign changes: [tex]\( + \to - \to + \to - \to + \to - \)[/tex]. There are 5 sign changes.
- The possible number of positive zeros is 5, 3, or 1 (subtract multiples of 2).
3. Determine the possible number of negative zeros:
- Evaluate [tex]\( f(-x) \)[/tex] instead of [tex]\( f(x) \)[/tex].
- This results in [tex]\( -25x^5 - 15x^4 - 145x^3 - 87x^2 - 100x - 60 \)[/tex].
- The signs remain constant as [tex]\( [-, -, -, -, -, -] \)[/tex] with no sign changes.
- Therefore, there are 0 possible negative zeros.
Summarizing:
- For [tex]\( f(x) = 4x^6 - 3x^4 - 16x^2 + 12 \)[/tex]:
- Possible imaginary zeros: 0, 2, 4, or 6
- Possible positive zeros: 3 or 1
- Possible negative zeros: 2 or 0
- For [tex]\( f(x) = 25x^5 - 15x^4 + 145x^3 - 87x^2 + 100x - 60 \)[/tex]:
- Possible imaginary zeros: 0, 2, or 4
- Possible positive zeros: 5, 3, or 1
- Possible negative zeros: 0
### For the function [tex]\( f(x) = 4x^6 - 3x^4 - 16x^2 + 12 \)[/tex]:
1. Determine the possible number of imaginary zeros:
- The degree of the polynomial is 6. Imaginary (or complex) zeros appear in conjugate pairs. Thus, the polynomial can have 0, 2, 4, or 6 imaginary zeros.
2. Determine the possible number of positive zeros:
- Use Descartes' Rule of Signs on [tex]\( f(x) \)[/tex].
- Look at the signs of the coefficients: [tex]\( [4, -3, -16, 12] \)[/tex] corresponds to signs [tex]\( [+, -, -, +] \)[/tex].
- Count the sign changes in the sequence: [tex]\( + \to - \to - \to + \)[/tex]. There are 3 sign changes.
- The possible number of positive zeros is therefore 3 or 1 (subtract multiples of 2 from the number of sign changes).
3. Determine the possible number of negative zeros:
- Evaluate [tex]\( f(-x) \)[/tex] instead of [tex]\( f(x) \)[/tex].
- [tex]\( f(-x) = 4x^6 - 3x^4 + 16x^2 + 12 \)[/tex].
- The signs are [tex]\( [+, -, +, +] \)[/tex].
- Count the sign changes: [tex]\( + \to - \to + \to + \)[/tex]. There are 2 sign changes.
- The possible number of negative zeros is 2 or 0 (subtract multiples of 2).
### For the function [tex]\( f(x) = 25x^5 - 15x^4 + 145x^3 - 87x^2 + 100x - 60 \)[/tex]:
1. Determine the possible number of imaginary zeros:
- The degree of the polynomial is 5. Imaginary zeros can again appear in pairs, so we can have 0, 2, or 4 imaginary zeros. However, we cannot have 6 here since the degree is not even.
2. Determine the possible number of positive zeros:
- Apply Descartes' Rule of Signs to [tex]\( f(x) \)[/tex].
- The sign sequence of the coefficients is [tex]\( [25, -15, 145, -87, 100, -60] \)[/tex], resulting in signs [tex]\( [+, -, +, -, +, -] \)[/tex].
- Count the sign changes: [tex]\( + \to - \to + \to - \to + \to - \)[/tex]. There are 5 sign changes.
- The possible number of positive zeros is 5, 3, or 1 (subtract multiples of 2).
3. Determine the possible number of negative zeros:
- Evaluate [tex]\( f(-x) \)[/tex] instead of [tex]\( f(x) \)[/tex].
- This results in [tex]\( -25x^5 - 15x^4 - 145x^3 - 87x^2 - 100x - 60 \)[/tex].
- The signs remain constant as [tex]\( [-, -, -, -, -, -] \)[/tex] with no sign changes.
- Therefore, there are 0 possible negative zeros.
Summarizing:
- For [tex]\( f(x) = 4x^6 - 3x^4 - 16x^2 + 12 \)[/tex]:
- Possible imaginary zeros: 0, 2, 4, or 6
- Possible positive zeros: 3 or 1
- Possible negative zeros: 2 or 0
- For [tex]\( f(x) = 25x^5 - 15x^4 + 145x^3 - 87x^2 + 100x - 60 \)[/tex]:
- Possible imaginary zeros: 0, 2, or 4
- Possible positive zeros: 5, 3, or 1
- Possible negative zeros: 0
