Factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$ How do you go about factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$
Which factoring method would be the best?

Notice that $t^4/4 = (t^2/2)^2$ so let $u = t^2/2$
Then $({16+4t^2+\frac{t^4}{4}})^{1/4} = (16 + 8u + u^2)^{1/4}$
$16 + 8u + u^2 = (4+u)^2$ and $(a^2)^{1/4} = a^{1/2}$
So $({16+4t^2+\frac{t^4}{4}})^{1/4} = (16 + 8u + u^2)^{1/4} = (4 +u)^{1/2} = (4+t^2/2)^{1/2}$
Is that simplified enough?
