To solve for the number of possible real zeros and turning points of the function [tex]\( f(x)=x^7-6x^6+8x^5 \)[/tex], we need to analyze the polynomial properties and determine the roots by factoring.
### Step-by-Step Solution:
1. Determine the Degree of the Polynomial:
The degree of the polynomial [tex]\( f(x) = x^7 - 6x^6 + 8x^5 \)[/tex] is 7, as the highest power of [tex]\( x \)[/tex] is [tex]\( x^7 \)[/tex].
2. Number of Possible Real Zeros:
According to the Fundamental Theorem of Algebra, a polynomial of degree [tex]\( n \)[/tex] can have at most [tex]\( n \)[/tex] real zeros. Therefore, a polynomial of degree 7 can have at most 7 real zeros.
3. Number of Turning Points:
The number of turning points of a polynomial is at most [tex]\( n - 1 \)[/tex], where [tex]\( n \)[/tex] is the degree of the polynomial. Here, the degree is 7, so the number of possible turning points is [tex]\( 7 - 1 = 6 \)[/tex].
4. Factoring the Polynomial:
To determine the real zeros, we factor the given polynomial [tex]\( f(x) = x^7 - 6x^6 + 8x^5 \)[/tex].
First, we factor out the greatest common factor (GCF):
[tex]\[
f(x) = x^5(x^2 - 6x + 8)
\][/tex]
Next, we factor the quadratic expression [tex]\( x^2 - 6x + 8 \)[/tex]:
[tex]\[
x^2 - 6x + 8 = (x - 2)(x - 4)
\][/tex]
Substituting this back into the polynomial, we get:
[tex]\[
f(x) = x^5(x - 2)(x - 4)
\][/tex]
5. Determine the Real Zeros:
The real zeros are found by setting each factor equal to zero:
[tex]\[
x^5 = 0 \Rightarrow x = 0
\][/tex]
[tex]\[
x - 2 = 0 \Rightarrow x = 2
\][/tex]
[tex]\[
x - 4 = 0 \Rightarrow x = 4
\][/tex]
Therefore, the real zeros of the polynomial [tex]\( f(x) \)[/tex] are [tex]\( x = 0 \)[/tex], [tex]\( x = 2 \)[/tex], and [tex]\( x = 4 \)[/tex].
### Summary:
- Number of Possible Real Zeros: Up to 7
- Number of Turning Points: Up to 6
- Real Zeros: [tex]\( x = 0 \)[/tex], [tex]\( x = 2 \)[/tex], and [tex]\( x = 4 \)[/tex]
Given the choices:
- 7 real zeros and 6 turning points; 0, 2, and 4
- 7 real zeros and 6 turning points; [tex]\( 0, -2 \)[/tex], and -4
- 6 real zeros and 5 turning points; 0, 2, and 4
- 6 real zeros and 5 turning points; [tex]\( 0, -2 \)[/tex], and -4
The correct one is:
6 real zeros and 5 turning points; 0, 2, and 4 (though it's a slight typo in the phrasing, it should be 3 real zeros for the exact problem given).
Therefore, the exact problem given has:
3 real zeros and 6 turning points; 0, 2, and 4.
