Solution:
The given polynomial is
f(x) = 6x^{3} 13x^{2}  4x + 15
Assume x = 1
f(1) = 6(1)^{3}  13(1)^{2}  4(1) + 15
f(1) = 6 13 + 4 + 15
f(1) = 0
Using the factor theorem, (x + 1) is a factor of the given polynomial.
6x^{3} 13x^{2}  4x + 15 / x + 1
=6x^{2}+ ( 19x^{2} 4x + 15 / x + 1)
= 6x^{2}+ ( (19x15) (x + 1) / (x + 1)
=6x^{2}  19x + 15
Dividing the polynomial by (x + 1) the quotient we get is 6x^{2}  19x + 15
6x^{3}  13x^{2}  4x + 15 = (x + 1)(6x^{2}  19x + 15)
= (x + 1)(6x^{2}  10x  9x + 15)
= (x + 1)[2x(3x  5)  3(3x  5)]
So we get,
= (x + 1)(2x  3)(3x  5)
Therefore, the completely factored form is (x + 1)(2x  3)(3x  5).
Summary:
The completely factored form of f(x) = 6x^{3} 13x^{2}  4x + 15 is (x + 1)(2x  3)(3x  5).
Impressive work there! The concept you've explored is polynomial factorization, specifically dealing with the factor theorem, polynomial division, and factoring techniques.
Let's break down the concepts used in the given article:

Polynomial: A mathematical expression consisting of variables and coefficients, involving addition, subtraction, multiplication, and nonnegative integer exponents. The given polynomial is (f(x) = 6x^3  13x^2  4x + 15).

Factor Theorem: It states that if (f(a) = 0), then (x  a) is a factor of the polynomial. In this case, when (x = 1), (f(1) = 0), which demonstrates that ((x + 1)) is a factor of (f(x)).

Polynomial Division: The process of dividing a polynomial by another polynomial. Here, (f(x)) is divided by ((x + 1)) to find the quotient, which is (6x^2  19x + 15).

Factoring: Expressing a polynomial as the product of its factors. The quadratic expression (6x^2  19x + 15) is factored further by splitting the middle term into ((2x  3)(3x  5)).

Completely Factored Form: The expression ((x + 1)(2x  3)(3x  5)) represents the completely factored form of the given polynomial (f(x) = 6x^3  13x^2  4x + 15).
This process showcases a strong understanding of polynomial factorization techniques, including the application of the factor theorem, polynomial division, and factoring by splitting the middle term in quadratics. The ability to decompose complex polynomials into their irreducible factors is crucial in various mathematical applications, including algebraic problemsolving and equationsolving tasks.