Factorise completely 6x^2 + x - 2. Hence, factorise completely 6(3m - 1)^2 + 3m - 3.

1. Split the middle term using two numbers with product $-12$ and sum $1$.

$6x^2+x-2=6x^2+4x-3x-2$

2. Factorise by grouping.

$6x^2+4x-3x-2=2x(3x+2)-1(3x+2)=(3x+2)(2x-1)$

1. Rewrite the expression so that it matches the quadratic from part (a).

Let $x=3m-1$.

Since $3m-3=(3m-1)-2$,
$6(3m-1)^2+3m-3=6x^2+x-2.$

2. Apply the factorisation from part (a) and simplify completely.

From part (a),
$6x^2+x-2=(3x+2)(2x-1).$

Substitute $x=3m-1$:
$[3(3m-1)+2][2(3m-1)-1]$
$=(9m-1)(6m-3)$
$=(9m-1)\cdot 3(2m-1).$

Alternatively, expanding and factorising directly gives
$6(3m-1)^2+3m-3=54m^2-21m=m(54m-21)=3m(18m-7).$

So the complete factorisation is
$3m(18m-7).$