What is the completely factored form of f(x)=6x^3-13x^2-4x+15 ? (x+1)(6x^2-19x+15) (x+1)^2(2x-3) (x+1)(3x-2)(5x-3) (x+1)(2x-3)(3x-5)

To find the completely factored form of the polynomial $f(x) = 6x^3 - 13x^2 - 4x + 15$, we can start by using the Rational Root Theorem to identify possible rational roots. The Rational Root Theorem states that any rational root of the polynomial is a factor of the constant term (15) divided by a factor of the leading coefficient (6).

The factors of 15 are $\pm 1, \pm 3, \pm 5, \pm 15$.
The factors of 6 are $\pm 1, \pm 2, \pm 3, \pm 6$.

Thus, the possible rational roots are:
$$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 3, \pm \frac{3}{2}, \pm 5, \pm \frac{5}{2}, \pm 15, \pm \frac{15}{2}$$

We can test these possible roots by substituting them into the polynomial to see if they yield zero.

Let's start with $x = 1$:
$$f(1) = 6(1)^3 - 13(1)^2 - 4(1) + 15 = 6 - 13 - 4 + 15 = 4$$
So, $x = 1$ is not a root.

Next, let's try $x = -1$:
$$f(-1) = 6(-1)^3 - 13(-1)^2 - 4(-1) + 15 = -6 - 13 + 4 + 15 = 0$$
So, $x = -1$ is a root.

Since $x = -1$ is a root, we can factor $f(x)$ as:
$$f(x) = (x + 1)(6x^2 - 19x + 15)$$

Now, we need to factor the quadratic $6x^2 - 19x + 15$. We can use the quadratic formula to find its roots:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 6$, $b = -19$, and $c = 15$.

$$x = \frac{19 \pm \sqrt{361 - 360}}{12} = \frac{19 \pm 1}{12}$$
$$x = \frac{20}{12} \text{ or } x = \frac{18}{12}$$
$$x = \frac{5}{3} \text{ or } x = \frac{3}{2}$$

So, the quadratic factors as:
$$6x^2 - 19x + 15 = 6(x - \frac{5}{3})(x - \frac{3}{2})$$

Thus, the completely factored form of $f(x)$ is:
$$f(x) = (x + 1)(6x^2 - 19x + 15)$$

Therefore, the correct answer is:
$$(x + 1)(6x^2 - 19x + 15)$$