Question 624035
Here's the trick:
Multiply the leading coefficient (6) times the independent term (12).
(Do not worry much about the sign of the product for now).
I got 72 as the product. How about you?
Look for factor pairs that multiply to give you that product, and list them.
I get:
(1)(72)=72
(2)(36)=72
(3)(24)=72
(4)(18)=72
(6)(12)=72
(8)(9)=72
and no other pair of factors.
 
OPTIONAL RATIONALE EXPLANATION:
If the factoring was {{{(ax+b)(cx+d)}}} ,
factors of that product, {{{ac*bd=ad*bc=6*12=72}}}, would appear as coefficients of terms in x before collecting like terms in the product of the factorization below.
{{{6x^2-17x+12=(ax+b)(cx+d)=highlight(acx^2+adx+bcx+bd)=acx^2+(ad+bc)x+bd}}} --> {{{ac*bd=6*12=72}}}
 
BACK TO THE TRICK (even if you do not understand or do not care for the rationale):
In the list of factor pairs above, look for factor pairs (consider all sign options now) that add up to the coefficient of the term in x (-17 in the case of the problem you posted) and multiply to give (6)(12)=72 (with its positive sign).
At this point, you find:
(-8)(-9)=72=(6)(12)
and (-8)+(-9)=-17
Then, -17x could be the result of -8x plus -9x, and you could write the trinomial to be factored as
{{{6x^2-17x+12=6x^2-8x-9x+12}}}
The expression on the right helps you figure out the factorization.
All you need now is some creative "taking out common factors" trick:
{{{6x^2-17x+12=6x^2-8x-9x+12=(6x^2-8x)+(-9x+12)=2x(3x-4)+(-3)(3x-4)=highlight((2x-3)(3x-4))}}}