Precalculus (6th Edition)

$\color{blue}{-\dfrac{4}{3}\sqrt{3}}$
Bring out the square root of $16$, which is 4, to obtain: $=-4\sqrt{\dfrac{1}{3}}$ Rationalize the denominator by multiplying $3$ to both the numerator and the denominator of the radicand to obtain: $=-4\sqrt{\dfrac{1(3)}{3(3)}} \\=-4\sqrt{\dfrac{3}{3^2}}$ Bring out the square root of the denominator to obtain: $=\color{blue}{-\dfrac{4}{3}\sqrt{3}}$