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The completely factored form of f(x) = 6x^{3} – 13x^{2} – 4x + 15 is (x + 1)(2x – 3)(3x – 5).

We know that after quadratic equations, we don’t have any direct method to find roots or factors. So for cubic and above equations we have to find roots by trial and error until we reach to quadratic equations.

So, Let us start with an assumption, say x = -1

Substituting x = -1 in the equation, we get

=> 6(-1)^{3} – 13(-1)^{2} – 4(-1) + 15

=> -6 -13 + 4 + 15 = 0

So, f(-1) = 0 which means -1 is a root and x + 1 is a factor of the given polynomial.

Dividing the polynomial by x+1, we get

=> 6x^{3} – 13x^{2} – 4x + 15 / (x + 1)

Splitting -13x^{2} as 6x^{2} – 19x^{2}

=> 6x^{3} + 6x^{2} + -19x^{2} – 4x + 15 / (x + 1)

Cancelling x+1

=> 6x^{2}(x+1) / (x+1) + (-19x^{2} – 4x + 15 / (x + 1))

=> 6x^{2 }+ (-19x^{2} – 4x + 15 / (x + 1))

Factorizing the other part and cancelling x+1

=> 6x^{2} + ( – (19x – 15) (x + 1) / (x + 1)

=> 6x^{2} – 19x + 15

So, the factorized form till now is as follows

6x^{3} – 13x^{2} – 4x + 15 = (x + 1)(6x^{2} – 19x + 15)

Now, we can directly factorize the quadratic factor

=> 6x^{2} – 19x + 15

=> 6x^{2} – 10x – 9x + 15

=> 2x(3x – 5) – 3(3x – 5)

=> (2x – 3)(3x – 5)

So the complete factored form of 6x^{3} – 13x^{2} – 4x + 15 is

=> (x + 1)(6x^{2} – 19x + 15)

=> (x + 1)(2x – 3)(3x – 5)