Question 920749
Try thinking of the quadratic trinomial this way.

{{{(ax+b)(cx+d)=12x^2-29+15}}}
{{{ac*x^2+bcx+adx+bd}}}
{{{ac*x^2+(bc+ad)x+bd}}}


Compare the coefficients to form the equations:
ac=12, -29=bc+ad, bd=15
Three equations but FOUR unknown variables.  You at least expect a,b,c,d to all be integers, and be nonzero.


You can make logical trials and test them.
a and c not relying on order nor sign, would be one of 1 & 12; 2 &6, 3 & 4. 
b and d similarly would be one of 1 & 15, or 3 & 5.


Do what you are able with those facts.


A more advanced way to factorize is to use the discriminant:
{{{(-29)^2-4*12*15=121}}},
{{{121=highlight(11^2)}}};
This means that the ROOTS for x are {{{(29-11)/24}}}  and  {{{(29+11)/24}}};
or, {{{18/24=3/4}}}  and  {{{40/24=8/3}}};
This then means you can choose a form for your quadratic expression,
{{{highlight(12(x-3/4)(x-8/3))}}}, which you could refactor if you want to.