Question 882684
{{{(x+u)(x+v)=x^2+(u+v)x+uv}}}
So,
{{{uv=12}}}
{{{u=12/v}}}
and,
{{{b=u+v}}}
{{{b=12/v+v}}}
So for any non-zero value of {{{v}}},
{{{b=12/v+v}}} and the equation is factorable to,
{{{x^2+(u+v)x+uv=(x+12/v)(x+v)}}}