Answer
$-(x+2)(x-2)(x^{2}+3)^{1/2}(x^{4}-x^{2}-13)$
Work Step by Step
We can factor out $(x^{2}-4)$, which is a difference of squares, and equals $(x+2)(x-2)$
Manipulating exponents, we can write $ A^{3/2}=A^{1+1/2}=A\cdot A^{1/2}$, meaning that we can factor out $(x^{2}+3)^{1/2}$ as well
Problem =$ (x+2)(x-2)(x^{2}+3)^{1/2}[1-(x^{2}-4)(x^{2}+3)]$
=$ (x+2)(x-2)(x^{2}+3)^{1/2}[1-x^{4}-3x^{2}+4x^{2}+12]$
=$ (x+2)(x-2)(x^{2}+3)^{1/2}[-x^{4}+x^{2}+13]$
Factor out -1 from the brackets, as the trinomial they hold is prime (we can't find two factors of 13 whose sum is 1).
= $-(x+2)(x-2)(x^{2}+3)^{1/2}(x^{4}-x^{2}-13)$
