Answer :
Final answer:
The possible rational zeros of the polynomial function [tex]\(f(x) = 11x^6 + 7x^3 + 7x^2 + 5x + 2\)[/tex] are -1/11, -2/11, -1, -2, 1, and 2.
Explanation:
Rational zeros of a polynomial are values of (x) that make the polynomial equal to zero when plugged in. In the given function [tex]\(f(x) = 11x^6 + 7x^3 + 7x^2 + 5x + 2\)[/tex], the coefficients are integers, so the rational zeros will be of the form [tex]\(\frac{p}{q}\)[/tex], where (p) is a factor of the constant term (2) and (q) is a factor of the leading coefficient (11).
The constant term's factors are [tex]\(\pm 1, \pm 2\)[/tex], while the leading coefficient's factors are [tex]\(\pm 1, \pm 11\)[/tex]. Therefore, we have four potential candidates for rational zeros: [tex]\(\pm \frac{1}{11}\) and \(\pm \frac{2}{11}\)[/tex].
Using the Rational Root Theorem, we can test these potential zeros by synthetic division. Plugging them into the function, we find that [tex]\(f(-\frac{1}{11}) = 0\) and \(f(-\frac{2}{11}) = 0\)[/tex]. These are two valid rational zeros.
However, the synthetic division shows that none of the other potential zeros [tex](\(\pm 1\) and \(\pm 2\))[/tex] are actual rational zeros for the function.
In summary, the possible rational zeros of the function [tex]\(f(x) = 11x^6 + 7x^3 + 7x^2 + 5x + 2\) are \(-\frac{1}{11}\), \(-\frac{2}{11}\)[/tex], -1, -2, 1, and 2.
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