Answer :
To use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros of the polynomial [tex]\( P(x) = -9x^4 - 2x^3 - x^2 - 3x + 6 \)[/tex], we follow these steps:
### (a) Possible number(s) of positive real zeros:
1. Identify the polynomial [tex]\( P(x) \)[/tex]:
[tex]\[ P(x) = -9x^4 - 2x^3 - x^2 - 3x + 6 \][/tex]
2. List the coefficients and their signs:
[tex]\[ -9, -2, -1, -3, 6 \][/tex]
3. Count the sign changes in the sequence of coefficients:
- The sequence is: [tex]\(-9\)[/tex] (negative), [tex]\(-2\)[/tex] (negative), [tex]\(-1\)[/tex] (negative), [tex]\(-3\)[/tex] (negative), [tex]\(6\)[/tex] (positive)
- There is one sign change, from [tex]\(-3\)[/tex] to [tex]\(6\)[/tex].
4. Apply Descartes' Rule of Signs:
- The number of positive real zeros is equal to the number of sign changes or less by an even number.
- Therefore, the possible number(s) of positive real zeros are:
[tex]\[ 1 \][/tex]
### (b) Possible number(s) of negative real zeros:
1. Substitute [tex]\( x \)[/tex] with [tex]\(-x\)[/tex] in [tex]\( P(x) \)[/tex] to get [tex]\( P(-x) \)[/tex]:
[tex]\[ P(-x) = -9(-x)^4 - 2(-x)^3 - (-x)^2 - 3(-x) + 6 \][/tex]
Simplify this expression:
[tex]\[ P(-x) = -9x^4 + 2x^3 - x^2 + 3x + 6 \][/tex]
2. List the coefficients and their signs of [tex]\( P(-x) \)[/tex]:
[tex]\[ -9, 2, -1, 3, 6 \][/tex]
3. Count the sign changes in the sequence of coefficients:
- The sequence is: [tex]\(-9\)[/tex] (negative), [tex]\(2\)[/tex] (positive), [tex]\(-1\)[/tex] (negative), [tex]\(3\)[/tex] (positive), [tex]\(6\)[/tex] (positive)
- There are three sign changes:
1. From [tex]\(-9\)[/tex] to [tex]\(2\)[/tex]
2. From [tex]\(2\)[/tex] to [tex]\(-1\)[/tex]
3. From [tex]\(-1\)[/tex] to [tex]\(3\)[/tex]
4. Apply Descartes' Rule of Signs:
- The number of negative real zeros is equal to the number of sign changes or less by an even number.
- Therefore, the possible number(s) of negative real zeros are:
[tex]\[ 3, 1 \][/tex]
So, summarizing:
(a) Possible number(s) of positive real zeros: [tex]\( 1 \)[/tex]
(b) Possible number(s) of negative real zeros: [tex]\( 3, 1 \)[/tex]
### (a) Possible number(s) of positive real zeros:
1. Identify the polynomial [tex]\( P(x) \)[/tex]:
[tex]\[ P(x) = -9x^4 - 2x^3 - x^2 - 3x + 6 \][/tex]
2. List the coefficients and their signs:
[tex]\[ -9, -2, -1, -3, 6 \][/tex]
3. Count the sign changes in the sequence of coefficients:
- The sequence is: [tex]\(-9\)[/tex] (negative), [tex]\(-2\)[/tex] (negative), [tex]\(-1\)[/tex] (negative), [tex]\(-3\)[/tex] (negative), [tex]\(6\)[/tex] (positive)
- There is one sign change, from [tex]\(-3\)[/tex] to [tex]\(6\)[/tex].
4. Apply Descartes' Rule of Signs:
- The number of positive real zeros is equal to the number of sign changes or less by an even number.
- Therefore, the possible number(s) of positive real zeros are:
[tex]\[ 1 \][/tex]
### (b) Possible number(s) of negative real zeros:
1. Substitute [tex]\( x \)[/tex] with [tex]\(-x\)[/tex] in [tex]\( P(x) \)[/tex] to get [tex]\( P(-x) \)[/tex]:
[tex]\[ P(-x) = -9(-x)^4 - 2(-x)^3 - (-x)^2 - 3(-x) + 6 \][/tex]
Simplify this expression:
[tex]\[ P(-x) = -9x^4 + 2x^3 - x^2 + 3x + 6 \][/tex]
2. List the coefficients and their signs of [tex]\( P(-x) \)[/tex]:
[tex]\[ -9, 2, -1, 3, 6 \][/tex]
3. Count the sign changes in the sequence of coefficients:
- The sequence is: [tex]\(-9\)[/tex] (negative), [tex]\(2\)[/tex] (positive), [tex]\(-1\)[/tex] (negative), [tex]\(3\)[/tex] (positive), [tex]\(6\)[/tex] (positive)
- There are three sign changes:
1. From [tex]\(-9\)[/tex] to [tex]\(2\)[/tex]
2. From [tex]\(2\)[/tex] to [tex]\(-1\)[/tex]
3. From [tex]\(-1\)[/tex] to [tex]\(3\)[/tex]
4. Apply Descartes' Rule of Signs:
- The number of negative real zeros is equal to the number of sign changes or less by an even number.
- Therefore, the possible number(s) of negative real zeros are:
[tex]\[ 3, 1 \][/tex]
So, summarizing:
(a) Possible number(s) of positive real zeros: [tex]\( 1 \)[/tex]
(b) Possible number(s) of negative real zeros: [tex]\( 3, 1 \)[/tex]