As mathematical operations, radicals (or roots) are the opposite of exponents.

In this tutorial, we'll learn all about multiplying radicals. But before we get started with that, here’s a quick recap of what radicals are.

Here’s what a radical looks like and what it consists of.

We read it as “the $\hspace{0.2em} 3 \hspace{0.2em}$^{rd} root of $\hspace{0.2em} 8 \hspace{0.2em}$″. Or the cube root (a special name for the $\hspace{0.2em} 3 \hspace{0.2em}$^{rd} root) of $\hspace{0.2em} 8 \hspace{0.2em}$.

As mathematical operations, radicals (or roots) are the opposite of exponents.

Let me explain. If we take $\hspace{0.2em} 2 \hspace{0.2em}$ and apply an exponent of $\hspace{0.2em} 3 \hspace{0.2em}$ to it, we get 8.

${\color{Orchid} 2} ^ {\color{Red} 3} \, = \, {\color{Teal} 8}$

Now how can we reverse this change and go back from $\hspace{0.2em} 8 \hspace{0.2em}$ to $\hspace{0.2em} 2 \hspace{0.2em}$? We take the third root of $\hspace{0.2em} 8 \hspace{0.2em}$.

$\sqrt[ {\color{Red} 3} ]{ {\color{Teal} 8} } \, = \, {\color{Orchid} 2}$

Similarly, $\hspace{0.2em} 3 \hspace{0.2em}$ raised to the second is $\hspace{0.2em} 9 \hspace{0.2em}$, so the second root of $\hspace{0.2em} 9 \hspace{0.2em}$ is $\hspace{0.2em} 3 \hspace{0.2em}$.

${\color{Orchid} 3} ^ {\color{Red} 2} \, = \, {\color{Teal} 9} \hspace{1em} \Leftrightarrow \hspace{1em} \sqrt[ {\color{Red} 2} ]{ {\color{Teal} 9} } \, = \, {\color{Orchid} 3}$

Makes sense?

Alright, now let’s see how we multiply radicals.

Note — If you don’t see an index on a radical, it means the index is 2. So,

$\sqrt{9} \, = \, \sqrt[ {\color{Red} 2} ]{9}$

To multiply radicals –

- Make sure each radical has the same index (if not, make the indices equal).
- Multiply the parts in front of the radicals (coefficients).
- Multiply the parts under the radical symbols (radicands).
- Multiply the parts under the radical symbols (radicands). Simplify the result.

If these steps don’t make much sense now, don’t worry. We’ll use them to solve several different examples in the following sections.

Example

Simplify: $\hspace{0.2em} \sqrt{5} \times \sqrt{6} \hspace{0.2em}$

Solution

Okay, this first one is very simple. Let’s go through the steps one by one.

Step 1. Make sure both radicals have the same index. They are both square roots (i.e. index of $\hspace{0.2em} 2 \hspace{0.2em}$).

$\sqrt{5} \times \sqrt{6}$

Step 2. There are no coefficients (parts in front of the radicals) to multiply, so the next step.

Step 3. Multiply the parts under the radical signs (radicands).

$\begin{align*} \sqrt{5} \times \sqrt{6} \, &= \, \sqrt{5 \times 6} \\[1em] &= \, \sqrt{30} \end{align*}$

Step 4. We can’t simplify it any further. $\hspace{0.2em} 30 \hspace{0.2em}$ doesn’t have any factors that are perfect squares (other than $\hspace{0.2em} 1 \hspace{0.2em}$, of course)

So $\hspace{0.2em} \sqrt{30} \hspace{0.2em}$ is our answer.

Example

Simplify: $\hspace{0.2em} \sqrt{6} \times \sqrt{15} \hspace{0.2em}$

Solution

Again, both are square roots and there are no coefficients to multiply. So, we can go ahead and multiply whatever is under the radical sign.

$\begin{align*} \sqrt{6} \times \sqrt{15} \, &= \, \sqrt{6 \times 15} \\[1em] &= \, \sqrt{90} \end{align*}$

But this time, there’s a little more we need to do. The result can be simplified further.

One of the factors of $\hspace{0.2em} 90 \hspace{0.2em}$ is $\hspace{0.2em} 9 \hspace{0.2em}$, which is a perfect square. So, we can take it out of the radical as $\hspace{0.2em} 3 \hspace{0.2em}$ ($\hspace{0.2em} 9 \hspace{0.2em}$ is $\hspace{0.2em} 3 \hspace{0.2em}$-squared).

$\begin{align*} \sqrt{90} \, &= \, \sqrt{ {\color{Red} 9} \times 10} \\[1em] &= \, \sqrt{ {\color{Red} 3^2} \times 10} \, = \, {\color{Red} 9} \sqrt{10} \end{align*}$

And that’s our answer.

Quick Tip

In most cases, it makes sense to factorize the radicands before you find their product. It makes simplification easier. Let me explain with an example.

$\begin{align*} \sqrt{35} \times \sqrt{21} \, &= \, \sqrt{35 \times 21} \\[1em] &= \, \sqrt{735} \end{align*}$

Here we multiplied $\hspace{0.2em} 35 \hspace{0.2em}$ and $\hspace{0.2em} 21 \hspace{0.2em}$ and got $\hspace{0.2em} 735 \hspace{0.2em}$. See the problem?

To simplify our answer, we’ll need to factorize $\hspace{0.2em} 735 \hspace{0.2em}$, looking for perfect square factors.

So why multiply $\hspace{0.2em} 35 \hspace{0.2em}$ and $\hspace{0.2em} 21 \hspace{0.2em}$ at all? Why not factorize them first? They are smaller compared to their product, $\hspace{0.2em} 735 \hspace{0.2em}$, so factorization would be easier. And we would save ourselves an unnecessary step of multiplication.

So, here's a smarter approach.

$\begin{align*} \sqrt{35} \times \sqrt{21} \, &= \, \sqrt{35 \times 21} \\[1em] &= \, \sqrt{(5 \times 7) \times (7 \times 3)} \end{align*}$

Factorizing at this step makes it clear that we have a pair of $\hspace{0.2em} 7 \hspace{0.2em}$s (or the square of $\hspace{0.2em} 7 \hspace{0.2em}$) in the factors and so we can take $\hspace{0.2em} 7 \hspace{0.2em}$ out of the radical sign to get the simplified and final answer.

$\begin{align*} \sqrt{(5 \times {\color{Red} 7} ) \times ( {\color{Red} 7} \times 3)} \, &= \, {\color{Red} 7} \sqrt{5 \times 3} \\[1em] &= {\color{Red} 7} \sqrt{15} \end{align*}$

As you will see in the examples below, the presence of coefficients doesn’t make much of a difference except that we need to multiply the coefficients together as well.

Example

Simplify: $\hspace{0.2em} 4 \sqrt{21} \times 5 \sqrt{30} \hspace{0.2em}$

Solution (a)

Step 1. Again, we start by making sure that both radicals have the same index. And again, they are both square roots. So, we can move to step 2.

$4 \sqrt{21} \times 5 \sqrt{30}$

Step 2. Time to multiply the coefficients, $\hspace{0.2em} 4 \hspace{0.2em}$ and $\hspace{0.2em} 5 \hspace{0.2em}$, together. So,

$\begin{align*} {\color{Red} 4} \sqrt{21} \times {\color{Red} 5} \sqrt{30} \, &= \, {\color{Red} 4} \times {\color{Red} 5} \times \sqrt{21} \times \sqrt{30} \\[1em] &= {\color{Red} 20} \times \sqrt{21} \times \sqrt{30} \end{align*}$

Step 3. Next, we multiply whatever is under the radicals.

${\color{Red} 20} \times \sqrt{21} \times \sqrt{30} \, = \, {\color{Red} 20} \times \sqrt{21 \times 30}$

Now, as I mentioned above, we don’t go ahead with the multiplication yet. Instead, we move to the next step.

Step 4. In this step, we simplify. So, let’s factorize and see if we can take something outside the radical.

$20 \times \sqrt{ {\color{Teal} 21} \times {\color{Orchid} 30} } \, = \, 20 \times \sqrt{ {\color{Teal} 3 \times 7} \times {\color{Orchid} 3 \times 10} }$

Ah, there’s one pair of $\hspace{0.2em} 3$s under the square root. That means we can replace them with one $\hspace{0.2em} 3 \hspace{0.2em}$ outside.

$\begin{align*} 20 \times \sqrt{ {\color{Red} 3} \times 7 \times {\color{Red} 3} \times 10} &= 20 \times {\color{Red} 3} \times \sqrt{7 \times 10} \\[1em] &= 60 \sqrt{70} \end{align*}$

And that’s it.

Example

Simplify: $\hspace{0.2em} 5 \sqrt{10} \times \sqrt{15} \hspace{0.2em}$

In this example, only the first radical seems to have a coefficient. But the thing to remember is – if you can’t see the coefficient, it’s $\hspace{0.2em} 1 \hspace{0.2em}$. So –

$5 \sqrt{10} \times \sqrt{15} \, = \, 5 \sqrt{10} \times {\color{Red} 1} \sqrt{15}$

But because multiplying or dividing by $\hspace{0.2em} 1 \hspace{0.2em}$ makes no difference, we might as well ignore it. Proceeding as we learned above –

$5 \sqrt{10} \times \sqrt{15} \, = \, 5 \sqrt{10 \times 15}$

And now we’ll factorize the numbers under the radical, looking for any pairs of factors that we can take out.

$\begin{align*} 5 \sqrt{10 \times 15} \, &= \, 5 \sqrt{2 \times {\color{Red} 5} \times 3 \times {\color{Red} 5} } \\[1em] &= \, 5 \times {\color{Red} 5} \sqrt{2 \times 3} \\[1em] &= \, 25 \sqrt{6} \end{align*}$

Done.

Example

Simplify: $\hspace{0.2em} \sqrt{3} \cdot (\sqrt{2} + 4 \sqrt{5}) \hspace{0.2em}$

Solution

Here, we have $\hspace{0.2em} \sqrt{3} \hspace{0.2em}$ being multiplied with a group of two terms. So, let’s distribute $\hspace{0.2em} \sqrt{3} \hspace{0.2em}$ across the parenthesis.

${\color{Red} 3} \cdot (\sqrt{2} + 4 \sqrt{5}) \, = \, {\color{Red} 3} \cdot \sqrt{2} + {\color{Red} 3} \cdot 4\sqrt{5}$

And now it looks much better. We just need to multiply each of the two pairs of radicals. We have done that many times now.

${\color{Teal} \sqrt{3} \cdot \sqrt{2}} + {\color{Orchid} \sqrt{3} \cdot 4 \sqrt{5}} \, = \, {\color{Teal} \sqrt{6}} + {\color{Orchid} 4 \sqrt{15}}$

We can't simplify it any further. So, this is our final answer.

Example

Simplify: $\hspace{0.2em} (\sqrt{2} + 4 \sqrt{3}) \cdot (5 \sqrt{2} + \sqrt{3}) \hspace{0.2em}$

Solution

In this example, we have two binomials multiplied together. Let’s expand the product using the FOIL method.

Of course, you can use any other method as well (matrix method?) if you are comfortable with it. The basic idea is that each term inside one pair of parenthesis gets multiplied with each term inside the other.

So —

And now we need to simplify each of the 4 terms.

Almost there. The two terms at the ends and the two in the middle are pairs of like terms. So we can add them and simplify the expression even more.

$\begin{align*} & {\color{Red} 10} + {\color{Teal} \sqrt{6}} + {\color{Teal} 20 \sqrt{6}} + {\color{Red} 12} \\[1em] = \,\, & {\color{Red} 22} + {\color{Teal} 21 \sqrt{6}} \end{align*}$

This is our answer.

Variables have not made an appearance in our examples so far. But we come across radicals with variables all the time. So let’s look at a couple of examples.

Example

Simplify: $\hspace{0.2em} 3x \sqrt{2x} \cdot 2y \sqrt{xy} \hspace{0.2em}$

Solution

As usual, we multiply the coefficients together and the radicands together.

$\begin{align*} 3x \sqrt{2x} \cdot 2y \sqrt{xy} \, &= \, 3x \cdot 2y \cdot \sqrt{2x} \cdot \sqrt{xy} \\[1em] &= \, 6xy \cdot \sqrt{2x \cdot xy} \end{align*}$

And now, we have a pair of $\hspace{0.2em} x$’s under the square root. So, we can take them out as $\hspace{0.2em} x \hspace{0.2em}$.

$\begin{align*} 6xy \cdot \sqrt{2 {\color{Red} x} \cdot {\color{Red} x} y} \, &= \, 6xy \cdot {\color{Red} x} \sqrt{2y} \\[1em] &= \, 6x^2y \sqrt{2y} \end{align*}$

That’s our answer.

Example

Simplify: $\hspace{0.2em} 4c \sqrt{2ab} \cdot 2 \sqrt{ab} \hspace{0.2em}$

Solution

Nothing too different here.

$\begin{align*} {\color{Red} 4c} \sqrt{2ab} \cdot {\color{Red} 2} \sqrt{ab} \, &= \, {\color{Red} 2} \cdot {\color{Red} 4c} \cdot \sqrt{2ab} \cdot \sqrt{ab} \\[1em] &= \, {\color{Red} 8c} \cdot \sqrt{2ab \cdot ab} \end{align*}$

There are two pairs of $\hspace{0.2em} ab \hspace{0.2em}$ under the square root. So we can replace them with one pair outside.

$\begin{align*} 8c \cdot \sqrt{2 {\color{Red} ab} \cdot {\color{Red} ab} } \, &= \, 8c \cdot {\color{Red} ab} \cdot \sqrt{2} \\[1em] &= \, 8 \sqrt{2} \, abc \end{align*}$

In all of the examples above, the radicals had the same index. But what about cases where radicals with different index numbers are to be multiplied?

Well, to multiply radicals with different indices, we must first make the indices equal. And for that, we’ll need to know how to change the index of a radical.

Here’s the key idea.

You can multiply the index of a radical with any number as long as you also multiply the exponents of everything under the radical by the same number.

Say you want to change the index of $\hspace{0.2em} \sqrt{15} \hspace{0.2em}$ from $\hspace{0.2em} 2 \hspace{0.2em}$ to $\hspace{0.2em} 8 \hspace{0.2em}$. No to turn $\hspace{0.2em} 2 \hspace{0.2em}$ into $\hspace{0.2em} 8 \hspace{0.2em}$, you must multiply it by $\hspace{0.2em} 4 \hspace{0.2em}$. So, here’s what you do.

Another example. Let’s change the index of cube root of ab squared into $\hspace{0.2em} 15 \hspace{0.2em}$. $\hspace{0.2em} 3 \hspace{0.2em}$ times $\hspace{0.2em} 5 \hspace{0.2em}$ is $\hspace{0.2em} 15 \hspace{0.2em}$, so we multiply by $\hspace{0.2em} 5 \hspace{0.2em}$.

$\begin{align*} \sqrt[3]{ab^2} \, &= \, \sqrt[3 {\color{Red} \times5} ]{a ^ {1 {\color{Red} \times 5} } \, b ^ {2 {\color{Red} \times 5} }} \\[1em] &= \sqrt[15]{a^5 \, b^{10}} \end{align*}$

Also, notice how we multiply the exponent of each variable by $\hspace{0.2em} 5 \hspace{0.2em}$.

Alright, time to dive into the real action.

Example

Simplify $\hspace{0.2em} \sqrt{2} \sqrt[3]{2} \hspace{0.2em}$ and write it using only one radical.

Solution

Step 1. When the radicals have different indices, start by taking their least common multiple (LCM) least common multiple (LCM). Here the indices are $\hspace{0.2em} 2 \hspace{0.2em}$ and $\hspace{0.2em} 3 \hspace{0.2em}$, so we take their LCM. The LCM is $\hspace{0.2em} 6 \hspace{0.2em}$.

Step 2. Next, using what we just learned, we make the index of each radical equal to the LCM, $\hspace{0.2em} 6 \hspace{0.2em}$.

Let’s do that for the first radical. Here, the index is $\hspace{0.2em} 2 \hspace{0.2em}$. So, we multiply by $\hspace{0.2em} 3 \hspace{0.2em}$.

Remember, invisible index means $\hspace{0.2em} 2 \hspace{0.2em}$. And invisible exponent means $\hspace{0.2em} 1 \hspace{0.2em}$.

$\begin{align*} \sqrt{2} \, &= \, \sqrt[2 {\color{Red} \times 3} ]{2^{1 {\color{Red} \times 3} }} \\[1em] &= \, \sqrt[6]{2^3} \end{align*}$

And then the next radical. $\hspace{0.2em} 3 \hspace{0.2em}$ times $\hspace{0.2em} 2 \hspace{0.2em}$ is $\hspace{0.2em} 6 \hspace{0.2em}$, so we multiply by $\hspace{0.2em} 3 \hspace{0.2em}$.

$\begin{align*} \sqrt[3]{2} \, &= \, \sqrt[3 {\color{Red} \times 2} ]{2^{1 {\color{Red} \times 2} }} \\[1em] &= \, \sqrt[6]{2^2} \end{align*}$

Step 3. Okay, now we have two radicals with the same index. So, the same old story – multiply what’s under the radicals, and simplify.

$\begin{align*} \sqrt[6]{2^3} \cdot \sqrt[6]{2^2} \, &= \, \sqrt[6]{2^3 \cdot 2^2} \\[1em] &= \, \sqrt[6]{2^5} \\[1em] &= \, \sqrt[6]{32} \end{align*}$

Just in case you’re wondering how $\hspace{0.2em} 2^3 \cdot 2^2 \hspace{0.2em}$ became $\hspace{0.2em} 2^5 \hspace{0.2em}$, when multiplying exponents with the same base ($\hspace{0.2em} 2 \hspace{0.2em}$, in our case here), we add the exponents.

Example

Simplify $\hspace{0.2em} \sqrt{a} \cdot \sqrt[4]{ab} \hspace{0.2em}$ and write it using only one radical.

Solution

In this example, the two indices are $\hspace{0.2em} 2 \hspace{0.2em}$ and $\hspace{0.2em} 4 \hspace{0.2em}$, and their LCM is $\hspace{0.2em} 4 \hspace{0.2em}$.

So, we need to change the index of the first radical to 4. We can leave the second one alone since its index is already $\hspace{0.2em} 4 \hspace{0.2em}$.

$\begin{align*} \sqrt{a} \, &= \, \sqrt[2 {\color{Red} \times 2} ]{a ^ {1 {\color{Red} \times 2} }} \\[1em] &= \, \sqrt[4]{a^2} \end{align*}$

And now we have two radicals with the same index, so we can multiply them easily.

$\begin{align*} \sqrt[4]{a^2} \cdot \sqrt[4]{ab} \, &= \, \sqrt[4]{a ^ 2 \cdot ab} \\[1em] &= \, \sqrt[4]{a^3 \, b} \end{align*}$

And with that, we come to the end of this tutorial on multiplying radicals. Until next time.

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