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radical_multiplication
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    Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. When multiplying radicals, the indices need to be the same and we literally just multiply the numbers inside and simplify if possible. So here we're going to get sqrt 8 * 11, but we know sqrt 8 we could think of as 4 * 2, and we know sqrt 4 is 2. So we would get 2 square roots of 22 as our final answer. If we look at another one, the indices are the same. They're both cube roots. So we're going to get the cube root of 21 seven times 3. We don't know the cube root of anything that can multiply to give us 21. Because I need 3 of something, I need 3 ^3 or 7 ^3. The indices are the same, so we just multiply the numbers on the inside. 6 * 318. The indices are the same again, it's the 5th root. 8 * 10 is 80 Y. To the 4th y ^3 * y we add the exponents. When the bases are the same, we don't know the 5th root of 80. We can think about 8 back here is 2 * 2 * 2, and 10 is 2 * 5 S The most twos we have are four of them, and we would need 5 to take one of them outside because the index is 5 sqrt y -, b times sqrt y + b. The indices are the same. So we're just going to multiply the radicans and we realize that's the difference of squares. Now we cannot take sqrt y ^2 -, sqrt b ^2 because y ^2 - b ^2 is all together one term. So that's our final answer there. This next one we just multiply the tops and we multiply the bottoms. So we get 7R over 5X here sqrt 28. We want to think about do we know a square that 28 is times So 4 * 7 is 28 and we know sqrt 4 is 2SO2 square roots of seven is our final answer 54. How about 6 * 9 is 54 sqrt 6 We don't know. But sqrt 9 is 3, so 3 square roots of 6 sqrt 50. How about 25 * 2? Sqrt 25 is 5 sqrt 2 we don't know. SO5 root 2 is our final answer. 175 Y to the 8th. Well, that one's a little bigger. We could do what's called a factor tree. We know that five's going to go into 175 because it ends in a zero or five S 5 into 17 is three times with two left over. SO35 into 25 S 5 * 35 is 175 and then 35 is 5 * 7. So sqrt 175 we could think of as 5 ^2 * 7 * y to the 8th. The square root and the square are going to cancel, so we're going to get a 5 sqrt 7. We don't know, so we're going to leave it as sqrt 7, sqrt y to the 8th. Well that square root is an understood two in the index, so we could think of it as Y to the 8 / 2 or Y to the 4th power. The fact that two went into 8 evenly means that Y to the 4th is on the outside. So our final answer there is going to be 5 Y to the 4th sqrt 7. This next one cube root of 54, well 54 is Even. So we can take a two out 2 * 27. 27 is 3 * 9 and 9 is 3 * 3. So I prime factored it. I know that the cube root of 54 is 2 * 3 ^3. I don't know the cube root of 2, but we do know the cube root of 3 ^3. The cube root and the cube are going to cancel, leaving us a three. So our final answer is going to be 3 cube roots of two 320. If we prime factor that, how about 10 and 32. SO two and five and 32 is 4 and 8. Bring down A2 bring down A5. Four turned into two times 28 is 2 * 4. Keep prime factoring them that 4 is 2 * 2. So we're going to get the cube root of 1234562 to the 6th times 5. This index here of three goes into the exponent of 6/2 full times. We could think of that as 2 to the 6th over three or two squared. The cube root of 5 we don't know, so our final answer would be 2 ^2 or four cube roots of five, The cube root of 27 AB to the 6th, while the cube root of 27 is 3. The cube root of A we don't know, and three goes into the exponent of 6/2 full times with none leftover. So 3B squared cube root of A this next one negative cube root and -32. First about the cube root of a negative. If this is an odd index, we can have a negative on the inside. The negative comes out for our final answer because the cube root of -1 is -132 would be 4 * 8 or 2 * 2 four times two 2 * 2 * 2 * 2 * 2. So we'd get the cube root of 2 to the 5th A to the 6th. 3 goes into 5. How many times? Well, three goes into the five, one full time with two leftovers. So we're going to have two squared that two in the exponent as my leftovers. 3 went into five one full time. 2 leftover three goes into the six two full times. So our final answer is going to be negative two a ^2 sqrt 4. Find the simplified form. Assume that X can be any real number. That's important. We're going to use absolute value brackets where necessary. So sqrt 81 is 9, sqrt X -, 1 ^2. Well, a square and a square root are going to cancel with the provision that that inside has to be positive. So we're going to put square root or absolute values around that X -, 1. You could if you wanted to have absolute value of 9 * X - 1. Those mean the same thing. This is the typical answer, but both of them are the same. Here we need to factor out A2. First we need to get it into a monomial. So we get X ^2 + 4 X +4. Now that actually factors into X + 2 quantity squared. We don't know sqrt 2, but we do know the square root of the quantity X + 2 ^2. Because it's an even index, we need to have the absolute value. We're going to assume that no raticans were formed by raising negative numbers to even powers. So now we don't care about the absolute value because we don't have negative numbers to start with. This is an understood index of two. The easiest way to do this is 6 / 2 is 3 and 2 into 9 goes four whole times with one leftover. The leftover always stays underneath the radical. SO2 goes into 6 evenly 3 full times. Two goes into 9/4 full times with one leftover. That's our answer here. Three goes into 6 twice, three goes into 7 twice with one leftover. Three goes into 13, four times with two leftover. That's our final answer. The 4th root of 16. Well, 16 we could think of as 2 * 2 * 2 * 2. So we're going to have four goes into four one time, four goes into 19 four times with three leftover, four goes into 13, three times with one leftover. So 2X to the 4th y ^3, the 4th root X ^3 y five goes into 13, twice with three leftover. Five goes into eight, once with three leftover. Five goes into 17, three times with two leftover. That's our final answer. Cube root of -80 We could think of -80. Let's first think of it as a -1 * 80, so we have -1 and then 88 * 10, so the cube root of -1. We always take the negative on the outside. If it's an odd index, 3 is odd, so it's an odd index. The cube root of 8 is 2. The cube root of 10 we don't know. And three goes into 14 four full times with two leftover. That's our final answer -2 eight of the fourth cube root, 10A squared. This next one we're going to multiply. So if I multiply 15 * 5, I could think of 15 as 3 * 5. So now hopefully we see we have 3 * 5 ^2 and sqrt 3. We don't know but sqrt 5 ^2 is just 5 S Our final answer is 5 sqrt 33. We could have also have done this as fifteen 5 * 5 is sqrt 75. Splitting 75 up into 3 * 25 sqrt 3 we don't know. Sqrt 25 is 5. This next one, the indices are the same, so we're going to multiply the radicands. The numbers inside 2 * 4 is 8, and the cube root of 8 is 2. Indices once again are the same, so we're going to multiply the radicands. 7 * 3 is 21 X times X ^2 is X ^3. We don't know the cube root of 21, but we do know the cube root of X ^3 being plain old X. This next one 5 * 15 we could do 5 * 3 * 5 A to the seventh times a cubed. When the bases are the same, we add the exponents. So if we rewrite this in simplified form, we'd get sqrt 3, which we don't know, Sqrt 5 ^2 the square root and the square are going to cancel, leaving us 5, and the square root of A to the 10th 2 goes into 10 five times with no leftovers. SO5A to the 5th sqrt 3 indices are the same again, so we're going to add the exponents when we multiply. 2 + 2 is 4/4 plus 6 is 10. Three goes into four one full time with one leftover. Three goes into 10 three full times with one leftover. The indices are the same, so we're going to multiply. When we have same basis, we add the exponents. The 4th root of 81. We can see that 81 is really 3 to the 4th power. So the 4th root of 3 to the 4th, X to the 9th Y to the 11th. The 4th root of 3 to the 4th is 3/4 goes into 9 two full times with one leftover. Four goes into 11/2 full times with three leftover. So our final answer, three X ^2 y ^2, the 4th root of XY cubed this one. The indices are the same. When the bases are being multiplied, we add the exponent, so we're going to get eight of the 10th b -, C to the 8th. Five goes into 10 two full times. Five goes into eight, one full time with three leftover. Thank you and have a wonderful day.