Question 816940
I assuming 2y^2+8y+6 

Since there is no real obvious factor we can use we should factor any common denominators in this equation (which is 2). So the equation is easier to work with. 

{{{ (2)(y^2+4y+3) }}}  

Accordingly, we have to factor out {{{ y^2+4y+3 }}} which will be: 

{{{(y+3)(y+1) }}}  

So in truth the facros we have are: 

{{{ (2)(y+3)(y+1) }}} 

or you could write it as: 

{{{(2y+6)(y+1)}}} 

and to check: 

F: {{{2y*y = 2y^2}}} 
O: {{{2y * 1 = 2y}}}
I: {{{6* y = 6y}}}
L: {{{6* 1 = 6}}} 

Which leaves us with our original equation: 

{{{ 2y^2 + 8y + 6}}}}