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 .

The second power and square root cancel out leaving you with just .

Let's factor a  out first since they divide evenly with all of the terms.

We get 

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 .

We have  perfect squares so the final answer is . 

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

  are perfect squares so the answer is . 

To simplify this expression, I like to think of it in terms of a factor tree. I simplify it down to its fully factored form and then for every pair of the same number, I place that number on the outside of the radical sign and cross out the pair.

The fully factored form is:

.

Then, when I account for the pairs that have been crossed out, the expression on the outside of the radical sign is:

.

Left on the inside of the radical sign is  

Then, put all of that together to get:

.

First, factor the coefficient:

It's definitely not obvious that 13 goes into 117 unless you examine the potential answers.

We can see that we have a group of two 2's, and a group of two 3's, so those will be on the outside of the radical. We also have 13 leftover.

Now we can look at the variables. We have which can easily be re-written as . This means that when we take the square root we'll get .

We have which can almost just as easily be written as , which is helpful because 90 is an even number. That can be re-written as , so when we take the square root we will get with one  leftover.

Putting this all together gives:

First, factor 2000. We don't have to break it down into prime factors before we encounter a clear cubic number:

Now we can look at the variables. Know that if you take the cube root of something to the third, you get that thing back. In other words, .

so when we take the cube root we get with one leftover.

so when we take the cube root we get with leftover.

so when we take the cube root we get and nothing is leftover.

Putting this all together, we get:

First, break 540 into its prime factors:

Its factors are . Within these factors we can recognize one pair of two 2's, and one pair of two 3's. That leaves a 3 and a 5 unmatched.

What this means is that we will have on the outside of the radical since we were able to take the square root of these pairs. Then we will have left on the inside since we can't take the square root of those.

Now we can look at the variables. Both have even powers:

so the square root of that would be . We can put on the outside of the radical.

so the square root is , and we can put that on the outside of the radical.

All of this together gives us or

First, factor -108:

We see that there is only one group of 3, the -3's

Note that this could be factored in a different way with respect to negatives:

If you did it this way, then there would still be a group of 3 3's and the negative would be inside the radical. That works, but conventionally the cube root of a negative number is negative, so it would seem to make more sense to move the negative outside the radical since we can.

Next, look at the variables. First, so the cube root of that is .

We can think about as . We can take the cube root of and get x, then have leftover.

Putting this all together, our answer is .

Simplifying gives .

To simplify this expression, I like to look at each term separately. First, I factor 72 down to its prime factors: . Since we're dealing with a square root, for every pair of the same number, we cross it out underneath the radical and keep the loners underneath the radical sign: . Do the same for all of the , , and  terms.  is the only one that has a loner underneath the radical. Multiply all of the terms to get your answer:

To simplify  we must break 375 into factors that we know the square root of. We do this because 375 is not a perfect square like 9 or 25. It is important to choose factors that have a perfect square so that we only deal with whole numbers.