Factorise 3x³+4x²-13x+6 completely, given that x-1 is a factor.

The coefficients of the polynomial $3x^3 + 4x^2 - 13x + 6$ are 3, 4, -13, and 6.

$$\begin{array}{c|cccc} 1 & 3 & 4 & -13 & 6 \\ & & 3 & 7 & -6 \\ \hline & 3 & 7 & -6 & 0 \\ \end{array}$$

The remainder is 0, confirming that $(x-1)$ is a factor. The quotient is $3x^2 + 7x - 6$.

We look for two numbers that multiply to $3 \times (-6) = -18$ and add up to 7. These numbers are 9 and -2.
Rewrite the middle term:

$$3x^2 + 9x - 2x - 6$$

Factor by grouping:

$$3x(x+3) - 2(x+3)$$ $$(3x-2)(x+3)$$

The complete factorization of the polynomial is the product of the given factor and the factors of the quadratic quotient:

$$(x-1)(x+3)(3x-2)$$

The factorization $(x-1)(x+3)(3x-2)$ matches option C.

The final answer is $\text{C}$