Question

add or subtract.
$4\sqrt{48} - 3\sqrt{32} + \sqrt{27}$
$4\sqrt{48} - 3\sqrt{32} + \sqrt{27} = \square$
(simplify your answer. type an exact answer, using radicals as needed.)

Explanation:

Step1: Simplify each radical term

$4\sqrt{48} = 4\sqrt{16\times3} = 4\times4\sqrt{3} = 16\sqrt{3}$
$-3\sqrt{32} = -3\sqrt{16\times2} = -3\times4\sqrt{2} = -12\sqrt{2}$
$\sqrt{27} = \sqrt{9\times3} = 3\sqrt{3}$

Step2: Combine like radical terms

$16\sqrt{3} + 3\sqrt{3} - 12\sqrt{2} = (16+3)\sqrt{3} - 12\sqrt{2}$

Answer:

$19\sqrt{3} - 12\sqrt{2}$