Question 133080
Well, that is not precisely correct.  ALL quadratic trinomials can be factored, just not necessarily to integers, rationals, or even real numbers.  In other words, for any second degree polynomial {{{ax^2+bx+c}}} there exist two numbers {{{p}}} and {{{q}}} such that {{{(x-p)(x-q)=ax^2+bx+c}}}.  It is just that in the case of your equation, {{{2x^2+4x+6=0}}}, {{{p}}} and {{{q}}} are a conjugate pair of complex numbers, namely {{{-1+-i*sqrt(2)}}}, where {{{i^2=-1}}}.


That means your factors would be {{{(x-(-1+i*sqrt(2)))(x-(-1-i*sqrt(2)))}}}


Having said all of that, I suspect you meant that the polynomial cannot be factored over the integers.  That is an absolutely correct statement.