To determine the possible number of positive and negative real zeros for the function [tex]\( f(x) = x^7 + x^6 + x^2 + x + 3 \)[/tex] using Descartes's Rule of Signs, follow these steps:
### Step 1: Determine the number of sign changes in [tex]\( f(x) \)[/tex]
To find the number of positive real zeros, examine the signs of the coefficients in the polynomial's terms:
[tex]\[
f(x) = x^7 + x^6 + x^2 + x + 3
\][/tex]
The coefficients are [1, 1, 0, 1, 1, 0, 1, 3]. Observe these coefficients:
- 1 (positive)
- 1 (positive)
- 0 (no change in sign)
- 1 (positive)
- 1 (positive)
- 0 (no change in sign)
- 1 (positive)
- 3 (positive)
Since there are no sign changes among the coefficients, there are 0 positive real zeros.
### Step 2: Determine the number of sign changes in [tex]\( f(-x) \)[/tex]
To find the number of negative real zeros, substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex] and then examine the signs of the coefficients:
[tex]\[
f(-x) = (-x)^7 + (-x)^6 + (-x)^2 + (-x) + 3 = -x^7 + x^6 + x^2 - x + 3
\][/tex]
So, the coefficients for [tex]\( f(-x) \)[/tex] are: [-1, 1, 0, -1, 1, 0, -1, 3]. Now, check the sign changes:
- -1 (negative)
- 1 (positive)
- 0 (no change in sign)
- -1 (negative)
- 1 (positive)
- 0 (no change in sign)
- -1 (negative)
- 3 (positive)
Count the sign changes:
1. From -1 to 1 (change of sign)
2. From 1 to -1 (change of sign)
3. From -1 to 1 (change of sign)
4. From 1 to -1 (change of sign)
5. From -1 to 3 (change of sign)
Thus, there are 5 sign changes in the coefficients of [tex]\( f(-x) \)[/tex].
### Step 3: Apply Descartes's Rule of Signs
According to Descartes's Rule of Signs:
- The number of positive real zeros is equal to the number of sign changes in [tex]\( f(x) \)[/tex] or less by an even integer.
- The number of negative real zeros is equal to the number of sign changes in [tex]\( f(-x) \)[/tex] or less by an even integer.
From our observations:
- There are 0 sign changes in [tex]\( f(x) \)[/tex], so there are 0 positive real zeros.
- There are 5 sign changes in [tex]\( f(-x) \)[/tex], which means the possible number of negative real zeros can be 5, 3, or 1 (since we subtract an even integer).
Out of the given answer choices, the correct one is:
D. 0 positive zeros, 3 or 1 negative zeros
