Answer
$\frac{ \sqrt 2}{t^{2}}$
Work Step by Step
$\frac{2 \sqrt 4 \sqrt 6}{\sqrt 16 \sqrt 3t^{4}}$
$\frac{2 \times 2 \sqrt 6}{4 \sqrt 3t^{4}}$
We simplify the constants $2 \times 2 \div 4 = 1$
$\frac{\sqrt 6}{\sqrt 3t^{4}}$
To simplify this, we cannot have a radical in the denominator.
We multiply $\frac{\sqrt 6}{\sqrt 3t^{4}}$ by $\frac{ \sqrt 3t^{4}}{ \sqrt 3t^{4}}$
$\frac{\sqrt 6}{\sqrt 3t^{4}} \times \frac{ \sqrt 3t^{4}}{ \sqrt 3t^{4}}$
$\frac{ \sqrt 18t^{4}}{ \sqrt (3t^{4})^{2}}$
Square root of $(3t^{4})^{2}$ is $3t^{4}$ because $3t^{4}$ x $3t^{4}$ = $(3t^{4})^{2}$
$\frac{ \sqrt 18t^{4}}{ 3t^{4}}$
The factors of $\sqrt 18t^{4}$ is $\sqrt 9t^{4} \times \sqrt 2$
$\frac{\sqrt 9t^{4} \times \sqrt 2}{ 3t^{4}}$
The square root of $\sqrt 9t^{4}$ is $3t^{2}$ because $3t^{2}$ x $3t^{2}$ =
$\sqrt 9t^{4}$
$\frac{ 3t^{2} \sqrt 2}{ 3t^{4}}$
$3t^{2} \div 3t^{4}$ = $t^{2}$
$\frac{ \sqrt 2}{t^{2}}$
