Question

PLEASE HELP PRECALCULUS WILL MARK BRAINLIEST SEE ATTACHMENT

Answer 1

Answer:

$\boxed{x = \pm \dfrac{1}{3}, \pm 1, \pm 3, \pm 9, \pm 27}$

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.

$\text{Possible root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

In your function, the constant term is 27 and the leading coefficient is 3, so

$\text{Possible root} = \dfrac{\text{factor of 27}}{\text{factor of 3}}$

Factors of 27 = ±1, ±3, ±9, ±27

 Factors of 3 = ±1, ±3

$\text{Possible roots are } \boxed{x = \pm \dfrac{1}{3}, \pm 1, \pm 3, \pm 9, \pm 27}$