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We learn how to multiply radicals using the product rule. Whenever we have the same index, we can simply multiply the radicals together. We can't multiply square roots and cube roots together but we could multiply fourth roots and fourth roots together since the indexes would match. Lets dive deeper into multiplying radicals and see if you can get some of these correct. See if you can multiply each pair of roots here together and then simplify the answer if possible. Try your best on these.

Here are the correct answers. Great work if you got those correct. For the first one we simply multiply the radicants together to get the square root of times 5, since the square root of 25 equals 5. We know 5 times 5 equals 25, our final answer. Another way to do this problem would be to multiply these two radicants together to get 625. So really we're taking the square root of 625 or whatever was underneath it in the first place, just 25. Now this is only true for square roots, it won't be true for high order roots. To start multiplying the third one, we want to multiply the numbers on the outside together first. So we'll have 5 times 5 on the outside and then the square root of 25 here. We know 5 times 5 equals 25 and we know the square root of 25 equals 5. We multiply one last time to get 125. This is the simplification for the fourth one, and we really just want to break down the 50 into a perfect square. We take the square root of 25, which equals 5, and then multiply these two numbers together to get 25 root 2. And finally for our last one, 2 times 12 equals 24. And then we multiply the cube roots together to get the cube root of 24, that's this 12 times 2. We split up 24 into 8 times 3 so that we can take the cube root of 8, which equals 2. We multiply one last time to get 48 times the cube root of 3. Nice effort on these problems.

## Multiply Radicals Practice 1

Let's try some more practice with the multiplying radicals. What do you think would be the result of these products? For the second one, you'll want to use the distributed property and multiply each term by 3 square root of 3. Write your final answers in the corresponding boxes, good luck.

## Multiply Radicals Practice 1

For our first problem, we'll have 24x squared and for our second problem we'll have 9 plus 3 times the square root of 6. Nice thinking for getting those two correct. For the first problem, we just want to multiply the radicands together. So we'll have 4 times 16, which equals 64, and x squared times x squared, which equals x to the 4th. The square root of 64 is 8 and the square root of x to the 4th is x squared. We want to make sure that we multiply these factors that come out by 3. So we'll have 24x squared for our final result. For the second problem we distribute or multiply 3 root 3 times each of each these terms. Here we'll multiply root 3 times root 3 to get 3. We still have a 3 in front of these two radicals, so we'll have 3 times 3, which equals 9. Then for the second term, we'll have 3 root 3, times root 2. We just multiply these two radicals together to get 3 root 6. There's a 1 in front of the square root of equal to 6. This one's underneath the root. We can't combine the 9 and the 3 root 6, since 9 doesn't have a square root of 6 on the end. These aren't like terms. So this is our final result.

## Multiply Radicals Practice 2

Try multiplying these radicals out. Keep in mind that you have a cubed root here. Now when we multiply it we don't change the index of the root. We just multiply by what's underneath. Then when we simplify, since we have a cubed root we only want to take out factors that appear three times underneath the radical. Use your knowledge of power divided by roots. And your knowledge of factors, to get to the simplified form.

## Multiply Radicals Practice 2

These are the correct answers, great work if you found those two. For the first one, we want to start by multiplying our numbers together, and our variables together. We'll have 20 times 2, which equals 40, x to the 5th times x squared which x to the 7th. The y squared times y to the 4th, which y to the 6th. Since under a cube root, we want to find factors that appear three times. We can split up 40 into 8 times 5 since the cubed root of 8 equals 2. We have to leave the 5 inside of our cube root since this factor doesn't appear 3 times. It's not a perfect cube. Finally we deal with our variables by using power divided by root. 7 divided by 3 is 2 remainder 1, so we'll have x squared on the outside and x to the 1 on the inside. The cubed root of y to the 6th, is just 6 divided by 3, so y squared comes out. This gives us our final result, I hope you found it. For the second one, we want to start by distributing 3 root 2 to each of these radical terms. Pull out 3 root 2 times root 18, and 3 root 2 times negative 2 root 6. For the first term we'll multiply these two radicals together to get 3 root 36. For our second term, we'll multiply 3 times negative square root of 36 equals 6, and then we can break down the square root of 12 into the square root of 4 times 3. We simplify further, by multiplying 3 times these two numbers together to give us our final answer, 18 minus 12 root 3. Nice work if you found that one as well.

## Multiply Binomials with Radicals 1

We've seen distribution with one radical, let's try multiply two binomials together. These binomials contain a radical in each one. This one has the square root of 2 and this one has the square root of 7. When we multiply two binomials, we want to distribute each term in the first parentheses to each term in the second parentheses. So we'll multiply 3 times 2 and 3 times root 7, and then we'll multiply root 2 times 2, and root 2 times root 7. We want to simplify any radicals that we can and combine like terms if possible. So try this one out, and don't worry if you don't get it right on your first try. We haven't even done one like this. If you think back to multiplying polynomials though, you just might figure it out. Good luck.

## Multiply Binomials with Radicals 1

We start by distributing the 3. So we'll have 3 times 2 for our first term. We'll add to that three 3 root 7 for our second term. Now that we've distributed the 3, let's distribute the root 2. We'll have root 2 times 2 for the third term, and root 2 times root 7 for our fourth term. Next, we multiply change anything since this is not inside of a radical. So we just have the 3 times the root of 7. Here we list the number and find the radical so we have 2 times root 2 and finally for this last term we have root 2 times root 7 which equals to root 14. We can't simplify any of these roots since, these numbers don't have factors that are perfect squares. Also none of these terms are like terms. This is just the number, this is root 7, this is root 2 and this is root together. So, we just get this lengthy sum of all the terms. That's our final result.

## Multiply Binomials with Radicals 2

Try multiplying this one out. We have the binomial of 3 root 3 plus 5 squared. Think about what the square really means, and then see if you can figure out the answer. As a caution, you can't distribute a square over a sum. We know that's not true. Again, think back to polynomials, and you might be able to figure it out. Try your best.

## Multiply Binomials with Radicals 2

Here's our final answer. Excellent work if you figured that one out. We know that a square or an exponent of two means that we write this out twice. We're multiplying this binomial by itself. Next we want to distribute 3 root 3 to our second binomial. So for the first term we'll have 3 root three times 3 root 3. For the second term we'll have 3 root 3 times 5. Once we've distributed that 3, the 3rd term and positive 5 times positive 5 for a 4th term, we multiply these two terms together to get 9 times 3, when were at the square root of any number times itself is just that number For these two terms, we'll 15 root 3, and 15 root 3, and on the end we have a positive 25. We can combine these terms together, since they both have a radical of square root of 3. So we have 30 square roots of 3, 9 times 3 is 27, and I move this 25 over here, so I can add my like terms together to get the final result. Great thinking if you got here.

## Multiply Binomials with Radicals 3

Try multiplying this third one out. You want to simplify the product of any radicals as you can, and then combine like terms as always. Good luck here.

## Multiply Binomials with Radicals 3

Our final answer is negative 19 plus the square root of 77. Nice work if you found this answer. Now this one was tough because we will end up distributing a square root of negative 11 in our problem. Let's see how it's done. First we distribute the square root of 7 to 2 root 7 and to 3 root 11. Next we distribute the negative square root of 11 to 2 root 7 and to 3 root 11... I write all these out so that way you don't confuse multiplying numbers and multiplying radicals. Like for this first term, I have 1 times 2 which equals times the square root of 77. For a third term, we'll have negative 2 times the square root of 77. And finally for our 4th term we'll have negative 3 times 11. The square root of 11 times the square root of 11 is just 11. We combine these two terms since they both have the square root of 77. 3 root 77 minus 2 root 77 is 1 root 77. We carry out the multiplication and then we'll have 14 minus 33, which equals negative 19. We bring down our root to get our final answer.

## Multiply Complex Numbers

we use a similar approach when multiplying context numbers. We just need to remember that i squared equals negative 1. So if ever we multiply two i's together, and we get i squared, we should replaced i squared with the value of negative 1. Essentially, it will just change the sign of a term that we're working with. So try using this knowledge on your own. See if you can multiply these two expressions together. Now I don't expect you to know this right off the bat, but if you use your knowledge of i squared equaling negative 1 and what you know about multiplication, I think you might get these. Either way have some fun and play around with the math.

## Multiply Complex Numbers

The first answer is negative 12 plus 8i, and the second answer is 22 plus 14i. If you figured either one of these out, great work. We start this first one by distributing the 4i to 2 and to 3i. 4i times 2 equals 8i, and 4i times 3i equals positive 12i squared. This is just kind of like x times x. We know x times x is x squared, so i times i is i squared. Okay, here comes the important part. We want to substitute i squared with the value of negative 1. So here's an i squared, so I'll replace it with negative 1. This is the key step to multiply complex numbers. You already know about distribution, and you already know about combining like terms, but you might not have known about that substitution. This is how we get a result of positive 12 plus 8i. For complex numbers, we usually list the real part first followed by the imaginary part. This is why I put the negative 12 here. We either have a plus bi, or we have a minus bi. These are the two forms that we'll see. For our second problem, we start by distributing this positive 3 to the 4 and to negative 2i. Next, we want to distrubte the 5i to our second parentheses. We'll have 5i times positive 4, and 5i time negative 2i. We can clean up our multiplication to get and 20i equals positive 14i. The other terms stay the same. Next, we have i squared. So we need to replace that with negative 1. This means we'll have negative 10 times negative 1, which would equal positive 10. Whenever we have an i square term it essentially makes the number negative. Like here, we had and we get positive 10. Now we're ready to combine our final like terms of 12 and 10 to get our final result. Now again, don't worry if you struggled with this one. You'll have a chance at the next one.

## Multiply Complex Numbers Practice 1

If you got both of the last one's correct, then now its time to show off your skills again. Otherwise if you might have stumbled the first time, now's your chance to shine. Try multiplying these expressions together, and then write your answer in the corresponding box below. Good luck.

## Multiply Complex Numbers Practice 1

For the first solution, we'll have negative 18 plus 24i. And for the second solution, we'll have 28 plus 16i. Nice work, if you got those two correct. We start by distributing the 6i to 4 and 3i. Next, we multiply to get 24i, and 18i squared. We replace i squared with its value of negative 1. So, essentially 18i squared becomes negative 18. Now we just flip the order of the real part and the imaginary part to get negative 18 plus 24i. In the second problem, we distribute positive 6 to 2 and to 4i. In the second distribution, we want to be careful. We have negative 4i times 2 here, and negative 4i times positive 4i here. We perform multiplication on each of our four terms to get 12 plus 24i minus 8i minus 16i squared. We combine our like terms of positive 24i and negative 8i to get positive 16i. Next, we want to replace this i squared with a value of negative 1. So negative 16i squared becomes positive 16. Remember, whenever we have an i squared term, the sign just changes, and we no longer have the variable of i squared. This was also true over here. We had positive one of those correct.

## Multiply Radicals Practice 3

Let's continue our practice with Multiplying Radicals with these two problems. What do you think these are in simplified form?

## Multiply Radicals Practice 3

The first answer is 3x, and our second answer is 2 plus 2 times the cube root of 6. Great thinking for getting those two correct. Now, these were a little bit tricky because you had to keep in mind we had cube roots. We can multiply these two radicals together since the indexes match, so we'll get 27x cubed. The factor of 3 appears three times underneath the radical as is the factor of x. This means we pull out one 3 and one x to get 3x. This should also make sense. We've seen the cube root of 27 before, and we know that's 3. And using power divided by root, we'll have 3 divided by 3 which is x to the 1. For our second problem, we'll distribute the cube root of 2 to each of these terms. So, we'll have the cube root of 8 and the cube root of 48. We know the cube root of times 6. We use our product rule to split this up into cube root of 8 times the cube root of 6, and simplify once again to get our final result. You might not have always shown this step throughout your problem solving, and that's okay. You just really want to be careful when you take cube roots of numbers. Put the factor on the outside, and leave anything else inside the radical.

## Multiply Radicals Practice 4

Here's our last problem for this lesson on multiplication on radicals. Try multiplying these two binomials out that contain cube roots. Think carfully when you simplify, and at the end, think about whether you can combine like terms or like radicals. Good luck.

## Multiply Radicals Practice 4

This multiplication, simplifies to this expression. Now this might look really complicated, but it's not too tough to get there. First, we distribute the cube root of 4xy to each of the terms in our second parentheses. When we multiply these two cube roots together, we'll get 64, x cubed, y cubed. So we know this distribution gives us this first term. Now we want to distribute this to the second term. For the second term, we'll have the cube root of 12, and then xy, from these variables here. Next we distribute the cube root of 2 to this term. So we'll have the cube root of 32x squared y squared. And finally for our last term we'll have the cubed root of 2 times the cubed root of 3. Which equals the cube root of 6. Next we try to simplify any of the radicals that we can. The cubed root of 64 equals 4. And the cubed root of x cubed y cubed is xy. This radical is already simplified since there's no factor that repeats 3 times underneath the radical, so we leave it as is. We can't simplify the x squared y squared since we don't have a group of 3. But we can simplify 32 here, since 32 is the same thing as 8 times 4. The cubed root of 8 equals 2, and this leads us to our answer. On the end here, the cubed root of 6 can't be simplified, so we have to tack it onto the end. There's no like terms here since none of the radicands are equal. So we're done. This is our final answer.