When doing this problem, we're going to simplify inside the radical first. So 1225, I know is divisible by 5, so I'm actually going to do a stair step method for division. So here's 55 goes into 12 twice with two leftover. So five goes into 22 four times with two leftover. Five goes into 25 five times. So 245 is again divisible by 55 goes into 24 four times. Five goes into 45 nine times. And then we can see 49 is 7 * 7. This is called the stair step method. And if you use this, you get the prime factorizations on the outside of the stair step. So that 1225 is going to be 5 ^2 * 7 ^2 and we have that AB to the 4th and we're going to still have to figure out that 1125 S 1125 because it ends in a 05. I know it is divisible by 55. Goes into 11 twice. One leftover five, goes into 12 twice with two leftover five goes into 25 five times. Once again, it ends in 05. So I know five again, 5 into 22 four times with two leftover. So 5 into 25545, we're only going to take out primes. So I'm going to take out a 5 again, and then nine is going to be three and three. So we're going to get 3 ^2 * 5 ^3 * A to the fifth times B. When I simplify this down, I'm going to see that the 5 ^2 and five cubed, there's one more on the bottom. So the leftovers go where there's more. So two on top, two on bottom, cancel seven squares up on top. There's an A on top and an A to the fifth on bottom. So that's going to leave me an A to the 4th on bottom. There's AB to the 4th on top and AB on bottom. That's going to leave me AB cubed on top. And I still haven't dealt with this 3 ^2, so I'm going to put the three squared in front. Now, there are a lot of different ways to do this, but one method would be to take the square root of things that we can do right now. So sqrt 7 ^2 of seven sqrt b ^3. This index of two goes into three one full time with one leftover. So the numerator is going to turn into seven B sqrt b sqrt 3 ^2 is going to give us 3 sqrt 5. We don't know. Square root of A to the 4th is a squared. Now we're wanting to rationalize the denominator so we can't have any radicals in the bottom. So now we're going to multiply by sqrt 5 / sqrt 5 in order to rationalize. So whenever we multiply by something over itself that's really just one, we multiply inside the radicals. So we get 7B square root of 5B over sqrt 5 times sqrt 5 is sqrt 25, which is really just five. So the denominator turns into 3 * 5 * a ^2, so fifteen a ^2.