Hello, wonderful mathematics people. This is Anna Cox from Kellogg Community College, Multiplying Radical Expressions. The first thing we want to make sure is that the indices are the same. If there's not a number there, it's understood that they're both 2. So we're going to multiply the numerators and the denominators. So this one's going to simplify to sqrt 7 R over 5X, and we can't do anything more with it right at this moment. The next one, they're both indices are two, so we're going to get sqrt 15 AB. The last one, the indices are both 3, so we're going to get the cube root of 21. We're going to let you try 4-5 and six on number six. You do not need to multiply it out. You can leave it as a monomial. So simplifying a radical by factoring what the square root means, the indicee of two means I need 2 of something to take it outside of the radical. So like if we asked for a sqrt 4, we would know that that's two, and sqrt 9 is 3 and sqrt 16 is 4, sqrt 25 is five, etcetera. Because all of these over here underneath the square root are really squares. We could think of four as 2 ^2. So sqrt 2 ^2, the square root in the square would cancel. Or sqrt 9 is sqrt 3 ^2 or sqrt 4 ^2. So the square root in the square cancel. So when I have 28, I'm going to think about what times what gives me 28? Well, 4 * 7. So I could think of this as sqrt 4 times sqrt 7 or 2 sqrt 7 when it's simplified. We could keep going further if we don't really understand the big picture here. And we could do sqrt 2 ^2 * 7 and then the index of two goes into the exponent of two one time. So two goes into two once. So that would mean that we bring 1-2 out and the two here can't go into the exponent of one evenly. So that would give us 27. Couple different ways to think about it. So 54 has an index of three, so 6 * 9 or 2 * 3 * 3 * 3. So if I think about having the cube root of 54 being the cube root of 2 * 3 ^3, I don't know the cube root of 2 because the exponent is 1, the index is 3, but the cube root of 3 ^3, the index and the exponent will cancel each other out. 3 / 3 is 1, so we're going to end up with three cube roots of two. This next one, 175, let's see 175. Five definitely goes in. Five goes into 17 three times. So 35557 to square root this time because there wasn't a number there in the indices. So we're going to think about the index of two divides into the exponent to one time for the five. The index of two doesn't go into one evenly, so we're going to leave the seven, and the index of two goes into the the exponent of eight four times evenly. So that's going to go on the outside. Now, if it hadn't been even, we put how many times evenly on the outside and the leftovers on the end. Here's a few for you to try. You know what, we should change that to a different number. Maybe I'll do that after my video. Find a simplified form of F of X. Assume that X can be any real number. Use absolute value brackets when necessary. So we know that the square root, if the index is even the inside has to be positive or zero. So whenever we have an even index, when we simplify it, we're going to have absolute values if there's a variable. If it's an odd index, we don't use absolute value, we get whatever it is out. So in 13, we're going to have the square root, we're going to pull out A2 and we're going to end up with X ^2 + 4 X +4, and that's going to factor into X + 2 * X + 2. So we could think of that as sqrt 2 times the quantity X + 2 ^2. Well, the index of two goes into the exponent of one, not evenly. So we're going to get sqrt 2 left alone. But this index of two goes into the exponent of two on this X + 2 one time. So we're going to use absolute values because we're bringing out a variable. So our solution is going to be the absolute value of X + 2 times sqrt 2. This next one we have 32. Well, that's 4 * 8 or 2 * 2 * 2 * 4 or 2222 and two. So the 5th root of 2 to the fifth X to the 16th, the index goes into the exponent one full time for the two. So it comes out five goes into 16 three times with one leftover. So we're going to bring out three full times, five goes into 16, three times for 15 with one leftover. Now, I'm not going to use absolute values because it was an odd index. So that's my final solution for that one. Let's look at the next page. Excuse me, There's a few for you to try. We're going to next grouping. We're going to simplify assume that no radicans were formed by raising negative exponents to even powers. So now we're not going to worry about if the inside is negative. We're just going to say the inside is always going to have variables that were positive. So the understood index is 2. Two goes into 6 three full times, Two goes into 9 four times, with one leftover. This next one we have a -16, so we're going to think of that actually as -1 * 16 we could think of as 4 * 4 or 2 * 2 * 2 * 2. Two to the 4th, A to the 6th, B to the 7th, C to the 13th. So we're going to take the index into each exponent. Now the cube root of -1 is really -1 because the negative times the negative times the negative, the cube root, the three goes into four one full time with one leftover. So we're going to leave 1/2 on the inside. We took 1-2 out, three goes into the 6th twice with none leftover. So a square comes out, three goes into 7 twice with one leftover and three goes in the 13 four times with one leftover. So we're going to get negative two a ^2 b ^2 C to the 4th cube root of 2 BC as our final answer this next 116 four times 4, two times two 2 * 2. So we're going to get 2 to the 4th, X to the 19th, Y to the 13th. The index of four is going to go into the exponent one full time. Four goes into 19 four times with three leftover. Four goes into 13 three times with one leftover. Thank you and have a wonderful day.