Answer :
To determine the number of possible positive and negative real zeros for the given function using Descartes' Rule of Signs, follow these steps:
### Given Function
The given polynomial function is:
[tex]\[ v(x) = 7x^6 - 3x^4 - 3x^3 - 7x^2 + 4x - 6 \][/tex]
### Part 1: Number of Possible Positive Real Zeros
Descartes' Rule of Signs states that the number of positive real zeros of a polynomial is either equal to the number of sign changes between consecutive non-zero coefficients or less than that by an even number.
1. Find the coefficients and observe the signs:
[tex]\[ 7, -3, -3, -7, 4, -6 \][/tex]
2. Count the number of sign changes:
- From [tex]\(7\)[/tex] to [tex]\(-3\)[/tex]: 1 sign change
- From [tex]\(-3\)[/tex] to [tex]\(-3\)[/tex]: no sign change
- From [tex]\(-3\)[/tex] to [tex]\(-7\)[/tex]: no sign change
- From [tex]\(-7\)[/tex] to [tex]\(4\)[/tex]: 1 sign change
- From [tex]\(4\)[/tex] to [tex]\(-6\)[/tex]: 1 sign change
Therefore, there are [tex]\(3\)[/tex] sign changes.
3. Possible number of positive real zeros:
According to Descartes' Rule of Signs, the number of possible positive real zeros is the number of sign changes or less by an even number. So, for [tex]\(3\)[/tex] sign changes, the possibilities are:
[tex]\[ 3,\, 1 \][/tex]
Thus, the number of possible positive real zeros is [tex]\(3\)[/tex] or [tex]\(1\)[/tex].
### Part 2: Number of Possible Negative Real Zeros
To determine the number of possible negative real zeros, consider [tex]\(v(-x)\)[/tex] and then apply the same rule.
1. Calculate [tex]\(v(-x)\)[/tex]:
[tex]\[ v(-x) = 7(-x)^6 - 3(-x)^4 - 3(-x)^3 - 7(-x)^2 + 4(-x) - 6 \][/tex]
Simplifying the exponents, we have:
[tex]\[ v(-x) = 7x^6 - 3x^4 + 3x^3 - 7x^2 - 4x - 6 \][/tex]
2. Find the coefficients and observe the signs:
[tex]\[ 7, -3, 3, -7, -4, -6 \][/tex]
3. Count the number of sign changes:
- From [tex]\(7\)[/tex] to [tex]\(-3\)[/tex]: 1 sign change
- From [tex]\(-3\)[/tex] to [tex]\(3\)[/tex]: 1 sign change
- From [tex]\(3\)[/tex] to [tex]\(-7\)[/tex]: 1 sign change
- From [tex]\(-7\)[/tex] to [tex]\(-4\)[/tex]: no sign change
- From [tex]\(-4\)[/tex] to [tex]\(-6\)[/tex]: no sign change
Therefore, there are [tex]\(3\)[/tex] sign changes.
4. Possible number of negative real zeros:
According to Descartes' Rule of Signs, the number of possible negative real zeros is the number of sign changes or less by an even number. So, for [tex]\(3\)[/tex] sign changes, the possibilities are:
[tex]\[ 3,\, 1 \][/tex]
Thus, the number of possible negative real zeros is [tex]\(3\)[/tex] or [tex]\(1\)[/tex].
### Summary
- Number of possible positive real zeros: [tex]\(3, 1\)[/tex]
- Number of possible negative real zeros: [tex]\(3, 1\)[/tex]
### Given Function
The given polynomial function is:
[tex]\[ v(x) = 7x^6 - 3x^4 - 3x^3 - 7x^2 + 4x - 6 \][/tex]
### Part 1: Number of Possible Positive Real Zeros
Descartes' Rule of Signs states that the number of positive real zeros of a polynomial is either equal to the number of sign changes between consecutive non-zero coefficients or less than that by an even number.
1. Find the coefficients and observe the signs:
[tex]\[ 7, -3, -3, -7, 4, -6 \][/tex]
2. Count the number of sign changes:
- From [tex]\(7\)[/tex] to [tex]\(-3\)[/tex]: 1 sign change
- From [tex]\(-3\)[/tex] to [tex]\(-3\)[/tex]: no sign change
- From [tex]\(-3\)[/tex] to [tex]\(-7\)[/tex]: no sign change
- From [tex]\(-7\)[/tex] to [tex]\(4\)[/tex]: 1 sign change
- From [tex]\(4\)[/tex] to [tex]\(-6\)[/tex]: 1 sign change
Therefore, there are [tex]\(3\)[/tex] sign changes.
3. Possible number of positive real zeros:
According to Descartes' Rule of Signs, the number of possible positive real zeros is the number of sign changes or less by an even number. So, for [tex]\(3\)[/tex] sign changes, the possibilities are:
[tex]\[ 3,\, 1 \][/tex]
Thus, the number of possible positive real zeros is [tex]\(3\)[/tex] or [tex]\(1\)[/tex].
### Part 2: Number of Possible Negative Real Zeros
To determine the number of possible negative real zeros, consider [tex]\(v(-x)\)[/tex] and then apply the same rule.
1. Calculate [tex]\(v(-x)\)[/tex]:
[tex]\[ v(-x) = 7(-x)^6 - 3(-x)^4 - 3(-x)^3 - 7(-x)^2 + 4(-x) - 6 \][/tex]
Simplifying the exponents, we have:
[tex]\[ v(-x) = 7x^6 - 3x^4 + 3x^3 - 7x^2 - 4x - 6 \][/tex]
2. Find the coefficients and observe the signs:
[tex]\[ 7, -3, 3, -7, -4, -6 \][/tex]
3. Count the number of sign changes:
- From [tex]\(7\)[/tex] to [tex]\(-3\)[/tex]: 1 sign change
- From [tex]\(-3\)[/tex] to [tex]\(3\)[/tex]: 1 sign change
- From [tex]\(3\)[/tex] to [tex]\(-7\)[/tex]: 1 sign change
- From [tex]\(-7\)[/tex] to [tex]\(-4\)[/tex]: no sign change
- From [tex]\(-4\)[/tex] to [tex]\(-6\)[/tex]: no sign change
Therefore, there are [tex]\(3\)[/tex] sign changes.
4. Possible number of negative real zeros:
According to Descartes' Rule of Signs, the number of possible negative real zeros is the number of sign changes or less by an even number. So, for [tex]\(3\)[/tex] sign changes, the possibilities are:
[tex]\[ 3,\, 1 \][/tex]
Thus, the number of possible negative real zeros is [tex]\(3\)[/tex] or [tex]\(1\)[/tex].
### Summary
- Number of possible positive real zeros: [tex]\(3, 1\)[/tex]
- Number of possible negative real zeros: [tex]\(3, 1\)[/tex]
