When simplifying radicals, you want to factor the radicand and look for square numbers.

\(\displaystyle \sqrt{6}=\sqrt{2\cdot 3}\) 

Both the \(\displaystyle 2\) and \(\displaystyle 3\) are not perfect squares, so the answer is just \(\displaystyle \sqrt{6}\). 

When simplifying radicals, you want to factor the radicand and look for square numbers.

\(\displaystyle \sqrt{36}=\sqrt{9\cdot 4}=3\cdot 2=6\)

 \(\displaystyle 36\) is a perfect square so the answer is just \(\displaystyle 6\). 

When simplifying radicals, you want to factor the radicand and look for square numbers.

\(\displaystyle \sqrt{729}=\sqrt{9\cdot81}=3\cdot9=27\)

If you aren't sure whether \(\displaystyle 729\) can be factored, use divisibility rules. Since it's odd and not ending in \(\displaystyle 5\), lets check if \(\displaystyle 3\) is a possible factor. To know if a number is factorable by \(\displaystyle 3\), you add the sum of the individual numbers in that number.

\(\displaystyle 729=7+2+9=18\)

Since \(\displaystyle 18\) is divisible by \(\displaystyle 3\), just divide \(\displaystyle 729\) by \(\displaystyle 3\) and continue doing this. You should do this \(\displaystyle 5\) more times and see that \(\displaystyle 729= 3\cdot3\cdot3\cdot3\cdot3\cdot3\). 

Since \(\displaystyle 9\) is a perfect square, or \(\displaystyle 3^2\) we can simplify the radical as follows.

\(\displaystyle \sqrt{729}=\sqrt{3\cdot3\cdot3\cdot3\cdot3\cdot3}=\sqrt{9\cdot9\cdot9}= 3\cdot3\cdot3=27\)

When simplifying radicals, you want to factor the radicand and look for square numbers.

\(\displaystyle \sqrt{8}=\sqrt{2\cdot 4}\)

 Since \(\displaystyle 4\) is a perfect square but \(\displaystyle 2\) is not, we can write it like this:

\(\displaystyle \sqrt{8}=\sqrt{2\cdot4}=2\sqrt{2}\) 

Remember, the other factor that's not a perfect square is left in the radicand and the square root of the perfect square is outside the radical. 

When simplifying radicals, you want to factor the radicand and look for square numbers.

\(\displaystyle \sqrt{600}=\sqrt{100\cdot6}\) 

Since \(\displaystyle 100\) is a perfect square but \(\displaystyle 6\) is not, we can write it as follows:

\(\displaystyle \sqrt{600}=\sqrt{100\cdot6}=10\sqrt{6}\) 

Remember, the other factor that's not a perfect square is left in the radicand and the square root of the perfect square is outside the radical. 

When simplifying radicals, you want to factor the radicand and look for square numbers.

\(\displaystyle \sqrt{242}=\sqrt{2\cdot 121}\) 

Since \(\displaystyle 121\) is a perfect square but \(\displaystyle 2\) is not, we can write it like this:

\(\displaystyle \sqrt{242}=\sqrt{2\cdot 121}=11\sqrt{2}\) 

Remember, the other factor that's not a perfect square is left in the radicand and the square root of the perfect square is outside the radical. 

If you didn't know \(\displaystyle 121\) was a perfect square, there's a divisibility rule for \(\displaystyle 11\). If the ones digit and hundreds digit adds up to the tens value, then it's divisibile by \(\displaystyle 11\). \(\displaystyle 1+1=2\) so \(\displaystyle 121\) is divisible by \(\displaystyle 11\) and its other factor is also \(\displaystyle 11\). It's best to memorize perfect squares up to \(\displaystyle 20.\)

When simplifying radicals, you want to factor the radicand and look for square numbers.

\(\displaystyle \sqrt{\frac{1}{4}}=\sqrt{\frac{1}{2}\cdot \frac{1}{2}}\) 

It's essentially the same factor so the answer is \(\displaystyle \frac{1}{2}\).

Let's see if this quadratic can be broken down. Remember, we need to find two terms that are factors of the c term that add up to the b term.

It turns out that \(\displaystyle x^2+4x+4 = (x+2)^2\).

The second power and square root cancel out leaving you with just \(\displaystyle x+2\).

Let's factor a \(\displaystyle 4\) out first since they divide evenly with all of the terms.

We get \(\displaystyle 4(x^2+2x+1)\)

Let's see if this quadratic can be broken down. Remember, we need to find two terms that are factors of the c term that add up to the b term.

It turns out that \(\displaystyle x^2+2x+1 = (x+1)^2\).

We have \(\displaystyle 2\) perfect squares so the final answer is \(\displaystyle 2(x+1)\). 

When simplifying radicals, you want to factor the radicand and look for perfect squares.

\(\displaystyle \sqrt{36x^2y^4}=\sqrt{9\cdot 4\cdot x\cdot x\cdot y^2\cdot y^2}=3\cdot2\cdotx\cdoty^2=6xy^2\)

 \(\displaystyle 36, x^2, y^4\) are perfect squares so the answer is \(\displaystyle 6xy^2\).