Answer :
To solve this problem, we need to determine the rational roots of the given polynomial function [tex]\( f(x) = x^6 - 2x^4 - 5x^2 + 6 \)[/tex].
### Step-by-Step Solution:
1. Identify the Polynomial Function:
[tex]\[ f(x) = x^6 - 2x^4 - 5x^2 + 6 \][/tex]
2. Rational Root Theorem:
According to the Rational Root Theorem, any rational root of the polynomial [tex]\( f(x) \)[/tex] in the form [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (6) and [tex]\( q \)[/tex] is a factor of the leading coefficient (1). Therefore, the potential rational roots are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 6 \][/tex]
3. Test the potential rational roots:
Evaluate [tex]\( f(x) \)[/tex] at each of these potential rational roots to check if they are actually roots.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^6 - 2 \cdot 1^4 - 5 \cdot 1^2 + 6 = 1 - 2 - 5 + 6 = 0 \][/tex]
Therefore, [tex]\( x = 1 \)[/tex] is a root.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^6 - 2 \cdot (-1)^4 - 5 \cdot (-1)^2 + 6 = 1 - 2 - 5 + 6 = 0 \][/tex]
Therefore, [tex]\( x = -1 \)[/tex] is a root.
The remaining potential rational roots ([tex]\( \pm 2, \pm 3, \pm 6 \)[/tex]) either:
- Do not satisfy [tex]\( x^6 - 2x^4 - 5x^2 + 6 = 0 \)[/tex], or
- Require more complex evaluation that does not yield zero.
4. Conclusion:
From the evaluation, we find that the only rational roots of the polynomial [tex]\( f(x) \)[/tex] are [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex].
Thus, there are [tex]\(\boxed{2}\)[/tex] rational roots for the polynomial [tex]\( f(x) = x^6 - 2x^4 - 5x^2 + 6 \)[/tex].
### Step-by-Step Solution:
1. Identify the Polynomial Function:
[tex]\[ f(x) = x^6 - 2x^4 - 5x^2 + 6 \][/tex]
2. Rational Root Theorem:
According to the Rational Root Theorem, any rational root of the polynomial [tex]\( f(x) \)[/tex] in the form [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (6) and [tex]\( q \)[/tex] is a factor of the leading coefficient (1). Therefore, the potential rational roots are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 6 \][/tex]
3. Test the potential rational roots:
Evaluate [tex]\( f(x) \)[/tex] at each of these potential rational roots to check if they are actually roots.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^6 - 2 \cdot 1^4 - 5 \cdot 1^2 + 6 = 1 - 2 - 5 + 6 = 0 \][/tex]
Therefore, [tex]\( x = 1 \)[/tex] is a root.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^6 - 2 \cdot (-1)^4 - 5 \cdot (-1)^2 + 6 = 1 - 2 - 5 + 6 = 0 \][/tex]
Therefore, [tex]\( x = -1 \)[/tex] is a root.
The remaining potential rational roots ([tex]\( \pm 2, \pm 3, \pm 6 \)[/tex]) either:
- Do not satisfy [tex]\( x^6 - 2x^4 - 5x^2 + 6 = 0 \)[/tex], or
- Require more complex evaluation that does not yield zero.
4. Conclusion:
From the evaluation, we find that the only rational roots of the polynomial [tex]\( f(x) \)[/tex] are [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex].
Thus, there are [tex]\(\boxed{2}\)[/tex] rational roots for the polynomial [tex]\( f(x) = x^6 - 2x^4 - 5x^2 + 6 \)[/tex].
