Question

multiply the radicals below. ensure your final answer is as simplified as possible:
$3\sqrt{21} \cdot -7\sqrt{12} = \square$

Explanation:

Step1: Multiply the coefficients and the radicands separately

First, multiply the coefficients \(3\) and \(-7\), and multiply the radicands \(\sqrt{21}\) and \(\sqrt{12}\) using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) (\(a\geq0,b\geq0\)). So we have \(3\times(-7)\times\sqrt{21\times12}\).

Step2: Calculate the product of coefficients and simplify the radicand

Calculate \(3\times(-7)= - 21\). Then simplify the radicand \(21\times12 = 252\). Now we need to simplify \(\sqrt{252}\). Factor \(252\) into prime factors: \(252=2\times126 = 2\times2\times63=2\times2\times3\times21 = 2\times2\times3\times3\times7\). So \(\sqrt{252}=\sqrt{2^{2}\times3^{2}\times7}\). Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) and \(\sqrt{a^{2}} = a\) (\(a\geq0\)), we get \(\sqrt{2^{2}\times3^{2}\times7}=\sqrt{2^{2}}\times\sqrt{3^{2}}\times\sqrt{7}=2\times3\times\sqrt{7} = 6\sqrt{7}\).

Step3: Combine the results

Now we have \(-21\times6\sqrt{7}\). Wait, no, wait. Wait, in step 1, we had \(3\times(-7)\times\sqrt{21\times12}\), and \(21\times12 = 252\), but let's re - check the simplification of \(\sqrt{252}\). Wait, \(21\times12=252 = 4\times63=4\times9\times7\). Oh, I made a mistake earlier. \(252 = 4\times9\times7\), so \(\sqrt{252}=\sqrt{4\times9\times7}=\sqrt{4}\times\sqrt{9}\times\sqrt{7}=2\times3\times\sqrt{7}=6\sqrt{7}\). Then the coefficient part: \(3\times(-7)=-21\). Then multiply \(-21\) with \(6\sqrt{7}\)? Wait, no, no. Wait, the original expression is \(3\sqrt{21}\cdot(-7\sqrt{12})\). So the correct way is:

\(3\sqrt{21}\times(-7\sqrt{12})=(3\times(-7))\times(\sqrt{21}\times\sqrt{12})\)

\(=-21\times\sqrt{21\times12}\)

\(=-21\times\sqrt{252}\)

Now, factor \(252\) to find perfect square factors: \(252 = 36\times7\) (because \(36\times7 = 252\), and \(36\) is a perfect square). So \(\sqrt{252}=\sqrt{36\times7}=\sqrt{36}\times\sqrt{7}=6\sqrt{7}\)

Then, \(-21\times6\sqrt{7}=-126\sqrt{7}\)

Wait, let's re - do the factoring of \(252\):

\(252\div4 = 63\), \(63\div9 = 7\), so \(252 = 4\times9\times7=2^{2}\times3^{2}\times7\), so \(\sqrt{252}=\sqrt{2^{2}\times3^{2}\times7}=2\times3\times\sqrt{7}=6\sqrt{7}\). Then \(3\times(-7)=-21\), and \(-21\times6\sqrt{7}=-126\sqrt{7}\)

Answer:

\(- 126\sqrt{7}\)