To find the possible rational zeros of the polynomial [tex]\( f(x) = -3x^4 - 9x^3 - 6x^2 - 8x + 14 \)[/tex], we can use the Rational Root Theorem. This theorem states that any possible rational root, in its lowest terms [tex]\( \frac{p}{q} \)[/tex], will have [tex]\( p \)[/tex] as a factor of the constant term (the term without [tex]\( x \)[/tex]) and [tex]\( q \)[/tex] as a factor of the leading coefficient (the coefficient of the term with the highest power of [tex]\( x \)[/tex]).
1. Identify the factors of the constant term (14):
The factors are: [tex]\( \pm 1, \pm 2, \pm 7, \pm 14 \)[/tex]
2. Identify the factors of the leading coefficient (-3):
The factors are: [tex]\( \pm 1, \pm 3 \)[/tex]
3. Form all possible rational numbers [tex]\( \frac{p}{q} \)[/tex] using these factors:
The possible rational numbers are obtained by dividing each factor of the constant term by each factor of the leading coefficient. This gives us:
- [tex]\( \frac{1}{1}, \frac{2}{1}, \frac{7}{1}, \frac{14}{1} \)[/tex]
- [tex]\( \frac{1}{3}, \frac{2}{3}, \frac{7}{3}, \frac{14}{3} \)[/tex]
- And their negatives [tex]\( -1, -2, -7, -14, -\frac{1}{3}, -\frac{2}{3}, -\frac{7}{3}, -\frac{14}{3} \)[/tex]
4. Combine and sort these possible zeros:
Arranging all these possibilities, we get:
[tex]\[
\pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}
\][/tex]
Thus, the list of all possible rational zeros of [tex]\( f(x) = -3x^4 - 9x^3 - 6x^2 - 8x + 14 \)[/tex] is:
[tex]\[
\pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}
\][/tex]
So, the correct answer is:
[tex]\[
\pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}
\][/tex]
This matches the third provided option in the question.
