Question 722966
Using only real number, you cannot factor {{{16x^2 + 81}}} ,
 
As you said
{{{16x^2-81=(4x+9)(4x-9)}}} can be factored, and when one or the other of those factors is zero (for x=9/4 or x=-9/4), their product is zero too, meaning {{{16x^2-81=0}}}.
 
{{{16x^2-72x+81=(4x-9)^2}}} or {{{16x^2+72x+81=(4x+9)^2}}} can be factored too, and become zero when they have a zero factor.
 
If you could factor {{{16x^2 + 81}}} using real numbers, you would be able to make it zero for some real value of x, and {{{16x^2 + 81=0}}} would have a solution.
However, for real values of x, {{{16x^2>=0}}} and {{{16x^2 + 81>=81}}} cannot be zero.
 
Maybe your daughter is studying complex numbers, or maybe she is supposed to say it cannot be factored, or maybe there was a typo.
 
You can factor {{{16x^2 + 81}}} using complex numbers, including the imaginary number {{{i}}} , whose square is {{{-1}}} 
{{{(4x-9i)(4x+9i)=16x^2-81*i^2=16x^2-81(-1)=16x^2+81}}}