Question 314498
{{{9x^4-44=(3x^2-sqrt(44))(3x^2+sqrt(44))}}}
Let me make a subsitution to make the factoring clearer.
{{{9x^4-44=(a^2x^2-b^2)(a^2x^2+b^2)}}}
where 
{{{a^2=3}}}
{{{a=sqrt(3) }}}
{{{b^2=sqrt(44)=2*sqrt(11) }}}
{{{b= sqrt(2*sqrt(11)) }}}
{{{9x^4-44=(ax-b)(ax+b)(ax-bi)(ax+bi)}}}
and {{{i=sqrt(-1)}}}
{{{9x^4-44=(sqrt(3)x-sqrt(2*sqrt(11))  )(sqrt(3)x+sqrt(2*sqrt(11))  )(sqrt(3)x-sqrt(2*sqrt(11)) i)(sqrt(3)x+sqrt(2*sqrt(11)) i)}}}