Answer:
[tex]\boxed{x = \pm \dfrac{1}{3}, \pm 1, \pm 3, \pm 9, \pm 27}[/tex]
Step-by-step explanation:
ƒ(x) = 3x³ + 39x² + 39x + 27
The Rational Zeros Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
[tex]\text{Possible root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}[/tex]
In your function, the constant term is 27 and the leading coefficient is 3, so
[tex]\text{Possible root} = \dfrac{\text{factor of 27}}{\text{factor of 3}}[/tex]
Factors of 27 = ±1, ±3, ±9, ±27
Factors of 3 = ±1, ±3
[tex]\text{Possible roots are } \boxed{x = \pm \dfrac{1}{3}, \pm 1, \pm 3, \pm 9, \pm 27}[/tex]
