Question

express in simplest radical form.
-8\sqrt{27} + 2\sqrt{108}

Explanation:

Step1: Simplify each radical

First, factor the radicands (the numbers inside the square roots) into perfect squares and other factors.
For \(\sqrt{27}\), we have \(27 = 9\times3\), and \(\sqrt{9\times3}=\sqrt{9}\times\sqrt{3}=3\sqrt{3}\) (since \(\sqrt{9} = 3\)).
For \(\sqrt{108}\), we have \(108 = 36\times3\), and \(\sqrt{36\times3}=\sqrt{36}\times\sqrt{3}=6\sqrt{3}\) (since \(\sqrt{36}=6\)).

Now substitute these back into the original expression:
\(-8\sqrt{27}+2\sqrt{108}=-8\times(3\sqrt{3}) + 2\times(6\sqrt{3})\)

Step2: Perform the multiplications

Calculate the products:
\(-8\times3\sqrt{3}=-24\sqrt{3}\)
\(2\times6\sqrt{3}=12\sqrt{3}\)

So the expression becomes:
\(-24\sqrt{3}+12\sqrt{3}\)

Step3: Combine like terms

Since both terms have \(\sqrt{3}\), we can combine the coefficients:
\((-24 + 12)\sqrt{3}=-12\sqrt{3}\)

Answer:

\(-12\sqrt{3}\)