Question 420164
{{{(5n^2+19n+12)}}}= multiply 5*12=60...Search among the factors of 60 to find two of them having sum 19 ( that is...4*15=60 4+15=19)
after that we have:
{{{(5n^2+15n+4n+12)}}}=
{{{(5n(n+3)+4(n+3))}}}=
{{{((n+3)(5n+4))}}}
etc.....
Same way:
{{{2v^2+11v+5 }}}=       We have 2*5=10 1*10=10 1+10=11
{{{2v^2+v+10v+5}}}=
{{{v(2v+1)+5(2v+1)}}}=
{{{(2v+1)(v+5)}}}

For a better way we use the {{{sqrt(b^2-4ac)}}} of the general formula {{{ax^2+bx+c}}} and if we find {{{b^2-4ac)0}}} then we cannot factorize.
I.e.the above:
{{{x^2-4x+24}}}
a=1,b=-4,c=24
and {{{b^2-4*1*24=(-4)^2-4*24=-80<0}}}