To solve for the number of rational roots of the polynomial function [tex]\( f(x) = x^6 - 2x^4 - 5x^2 + 6 \)[/tex], we follow these steps:
1. Finding the roots of the function: We need to determine the roots of the polynomial [tex]\( f(x) \)[/tex]. The roots are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].
2. Listing the roots: The roots of [tex]\( f(x) \)[/tex] are:
[tex]\[
x = -1, \quad x = 1, \quad x = -\sqrt{3}, \quad x = \sqrt{3}, \quad x = -\sqrt{2} \cdot i, \quad x = \sqrt{2} \cdot i.
\][/tex]
3. Identifying rational roots: A rational number is any number that can be expressed as a ratio of two integers [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are integers, and [tex]\( q \neq 0 \)[/tex]. The rational roots from the list given are:
[tex]\[
x = -1 \quad \text{and} \quad x = 1.
\][/tex]
The other roots, [tex]\( x = -\sqrt{3} \)[/tex], [tex]\( x = \sqrt{3} \)[/tex], [tex]\( x = -\sqrt{2} \cdot i \)[/tex], and [tex]\( x = \sqrt{2} \cdot i \)[/tex], are not rational. [tex]\( \sqrt{3} \)[/tex] is an irrational number, and [tex]\( \sqrt{2} \cdot i \)[/tex] represents imaginary numbers, which are not considered rational.
4. Counting the rational roots: Summarizing the results, we find that there are a total of 2 rational roots, which are [tex]\( -1 \)[/tex] and [tex]\( 1 \)[/tex].
Thus, the number of rational roots of the polynomial function [tex]\( f(x) = x^6 - 2x^4 - 5x^2 + 6 \)[/tex] is [tex]\( \boxed{2} \)[/tex].
