Question 114249
{{{12w^2 + 19w + 4=(aw+b)(cw+d)}}}
Find a,b,c,d using the FOIL method (First, Outer, Inner, Last) and equating to the coefficients of your equation.

Now to expand the equation you use the FOIL method (First, Outer, Inner, Last)
{{{(aw+b)(cw+d)}}}
First : {{{(highlight(aw)+b)(highlight(cw)+(d))=(aw)(cw)=acw^2}}}
Outer : {{{(highlight(aw)+b)(cw+highlight(d))=(aw)(d)=adw}}}
Inner : {{{(aw+highlight(b))(highlight(cw)+(d))=(b)(cw)=bcw}}}
Last : {{{(aw+highlight(b))(cw+highlight(d))=(b)(d)=bd}}}
{{{(aw+b)(cw+d)=(ac)*w^2+(ad)*w+(bc)w+bd}}}
{{{(aw+b)(cw+d)=(ac)*w^2+(ad+bc)*w+bd}}}
Comparing to your equation 
{{{12w^2 + 19w + 4=(ac)*w^2+(ad+bc)*w+bd}}}
1.{{{ac=12}}}
2.{{{ad+bc=19}}}
3.{{{bd=4}}}
From 1, since ac=12, you can choose 12 and 1, 6 and 2, 4 and 3, and also 3 and 4, 2 and 6, and 1 and 12 for possible a-c combinations.
From 3, since bd=4, you can choose 4 and 1, 2 and 2, and 1 and 4 for possible b-d combinations.
Let's start with a=6 and c=2
From 2, 
{{{6d+2b=19}}} 
Notice the coefficients of b and d. 
This solution set cannot be a solution because 2 even numbers will never sum to an odd number. 
So 6 and 2 and 2 and 6 are out as possible a-c combinations. 
Start again,
Let's choose a=12 and c=1, 
From 2, 
{{{12*d+b=19}}}
Now substitute possible b,d values.
{{{12*1+4=16}}} for b=4, d=1.  
{{{12*2+2=26}}} for b=2, d=2.
{{{12*4+1=49}}} for b=1, d=4.
No solutions for a=12 and c=1. 
Our last chance is 4 and 3.
Let's start with a=4 and c=3.
From 2, 
{{{4*d+3*b=19}}}
Again, substitute possible b,d values.
{{{4*1+3*4=16}}} for b=4, d=1  
{{{4*2+3*2=14}}} for b=2, d=2 
{{{4*4+3*1=19}}} for b=1, d=4 
Looks like a winner. 
a=4,c=3,b=1,d=4
{{{12w^2 + 19w + 4=(4w+1)(3w+4)}}}