Section 10.3: Simplifying Radical Expressions
§1 Use The Product Rule For Radicals
We can only multiply radicals if the indexes are the same. The product rule for radicals says that n a  n b  n ab .
You must remember that we can only multiply the radicals if the indexes are the same. For example, something
like
5  7  35 , but 5  3 7 cannot be combined.
The quotient rule says the same thing, except we are dividing radicals. Hence
n
a na

. Remember, if you
b nb
can simplify the radical, then you must always do so!
3
8
8
3
. However, this is not the final answer because the radicals are both perfect cubes.
125
125
2
Hence the solution is
.
5
For example,
3
§2 Simplify Radicals
Note that we can simplify 16 , which is 4, and also something like 3 27 which is 3. That’s because 16 is a
perfect square and 27 is a perfect cube. What happens when the radical is not a perfect square? For example,
what would the simplified form of
20 be?
There are four conditions for a radical to be simplified. First, the radicand has no factor raised to a power greater
than or equal to the index. Second, the radicand has no fractions. Third, no denominator contains a radical. And
last, exponents in the radicand and the index of the radicand have a GCF of 1.
So what does all this mean?
Basically, if the index is two, then you never want to leave a factor of the radicand that is a perfect square. If the
index is three, you never want to leave a factor in the radicand that is a perfect cube, etc… Also, we never leave
a radical in the denominator. We use a process called rationalization to properly simplify radicals in the
denominator.
When simplifying radicals, you need to check if a perfect square is a factor of the radicand. If it is, you can ‘take
it out’ of the radical using the product rule. For example, we know that
say that
20  4  5 . Of course, we can also
20  2 10 , but we note that neither is a perfect square. So the next step would be to use the
product rule to separate the radicals -
20  4  5  4  5  2 5 . Hence this is the final answer! Note that
the radicand is 5, and does not have a factor that is a perfect square.
Now try to simplify
down
80 . What is the greatest perfect square that is a factor of 80? Note that we can break
80  4  20 , but there is a larger perfect square that is a factor of 80. We can do
80  16  5  4 5 .
Remember, you need to keep track of the index as well. We can do 3 48  3 8  6  2 3 6 . Make sure you break
these down so that you don’t make any mistakes. Hopefully the more you do these types of problems it will be
easier to simplify. That’s why I always recommend memorizing the first 15 perfect squares and the first 6 perfect
cubes.
PRACTICE
1) Simplify
2) Simplify
200 and 108
3
128 and 3 96
§2 Simplify Radicals With Variables
Let’s look at an example first. Say we want to simplify
28x 4 y 7 . Note that we can use the product rule to split
up the radical into three parts. Essentially, we are trying to find the square root of three different terms here.
28 should be easy. How do we deal with the variables?
Well, there is a long way and there is a little shortcut. Let’s look at the long way first. We know that
x 4  x  x  x  x . Remember, the square root is asking you to find a perfect square in the radicand, because
then we can ‘take it out’. Also remember that x  x . Hence we see that really we are ‘taking out’ the x’s in
pairs. How many pairs of x’s are in the radicand? There are two. Hence we can take out the two pairs. Hence
2
x 4  x 2  x 2  x  x  x 2 . Similarly,
y 7  y  y  y  y  y  y  y  y 2  y 2  y 2  y . Hence we see that there
are three pairs of y 2 in the radicand. Then
y 7  y 3 y . So the final answer
28 x 4 y 7  2 x 2 y 3 7 y .
The shortcut simply says, divide the exponent of the variable by the index. The quotient tells you how many of
the variable comes out – the remainder tells you how many of the variable stays inside the radicand. If we look
at the previous example again,
28x 4 y 7 , we see that since the index is 2, we can divide each variable exponent
by 2. The result of the division is the exponent of the variable that comes outside. Any remainder from the
division is the exponent of the variable that stays inside. Hence, for x 4 , we divide the 4 by 2 to get 2, with a
remainder of zero. Hence x 2 comes out, and there is no x in the radicand. Similarly, for y 7 , we divide 7 by 2 to
get three, with a remainder of 1. Hence, y 3 comes out of the radical and one y remains inside. So the solution is
2x2 y3 7 y .
PRACTICE
40x5 yz 6
3) Simplify
4) Simplify
3
64x 2 y 8
§3 Simplify Products And Quotients Of Radicals With Different Indexes
Previously, we mentioned that something like 5  3 7 cannot be multiplied together because the indexes are
different. Technically, this is true. However, if we convert to rational exponents, we see that we can convert the
exponents so that they have the same index. For example, 5  3 7  51 2  71 3 . Note that the denominators of
the exponents are different. Can we make them the same? Yes we can! We can get the LCD and build the
equivalent fraction. Hence 51 2  71 3  53 6  72 6 . Now we can convert back to exponential form to get
51 2  71 3  53 6  7 2 6  6 53  6 7 2 . Now since they have the same index, we can multiply them together. The final
answer becomes
6
53  7 2 (you can leave the answer like this).
So remember, if the index is different, we need to first convert from radical form to exponential form. Then we
can compare the indexes and build the equivalent fraction so that they have the same index. Then, we convert
back to radical form and use the product rule to multiply.
PRACTICE
5) Simplify
5 3 2