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Multiplying and dividing radicals makes use of the "Product Rule" and the "Quotient Rule" as seen at the right. The "n" simply means that the index could be any value. The examples on this page use square and cube roots.
statement
Multiplying Radicals
When multiplying radicals (with the same index), multiply under the radicals, and then multiply any values directly in front of the radicals.

expin1
mu math1
ANSWER: mu math1c
 

Multiply the values under the radicals. mu math1a
Then simplify the result. mu math1b

 

expin2
mu math2
ANSWER: mu math2c
 
Multiply out front and multiply under the radicals.mu math2a
Then simplify the result.
mu math 2b


Product Rule

rule n multwhere a ≥ 0, b≥ 0

"The radical of a product is equal to the product of the radicals of each factor."


Quotient Rule

rule n div
where a ≥ 0, b > 0

"The radical of a quotient is equal to the quotient of the radicals of the numerator and denominator."

expin3
mu math3
ANSWER: mu math3c
 
Multiply under the radicals. mm3aa
Then simplify the result.
mu math3b


expin4
mu math4
ANSWER: mu math4c
 
Distribute across the parentheses. Remember there is an implied "1" in front of 4aa.
mu math4a
Then simplify the result. mu math4b


expin5
d
ANSWER: radex5c
 
Use the distributive property to multiply. Combine like terms.
radex5a


expin6
radex6
ANSWER: dnranex6aans
 
Use the distributive property to multiply. There are NO like terms to be combined.
dmradex6a


expin7 mdv4 ANSWER: mdv7
 

In this problem, it is easier to reduce the radicals before multiplying since the perfect cube (27) can be more clearly seen in each radicand. Yes, you could have chosen, instead, to multiply and then reduce.
mdv5


tidbit
A conjugate is a binomial formed by negating the second term of a binomial.
Example: (x + y) and (x - y) are conjugates.
A conjugate involving an imaginary number is called a complex conjugate.
Example: (a + bi) and (a - bi) are complex conjugates


expin8 md1 ANSWER: x2 - 3
  These terms are conjugates involving a radical. As with all conjugates, when multiplied, the middle terms cancel each other out. Notice the squaring of the square root. Notice that when the conjugates were multiplied, the radicals disappeared!
md2


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statement
Dividing Radicals
When dividing radicals (with the same index), divide under the radical, and then divide the values directly in front of the radical.

mu math5
ANSWER: mu math5c
 
Divide out front and divide under the radicals. mu math5a
Then simplify the result. mu math5b


expin2
mu math6
ANSWER: mumath6c
 

This fraction will be in simplified form when the radical is removed from the denominator. If we create a perfect square under the square root radical in the denominator the radical can be removed. To do so, we multiply the top and bottom of the fraction by the same value (this is actually multiplying by "1"). If we multiply by the square root radical we are trying to remove (in this case multiply by rad5), we will have removed the radical from the denominator.
mumath6a You have just "rationalized" the denominator!
You turned an irrational value into a rational value in the denominator.


expin3 mdv9 ANSWER: mdv11
 

We need to "rationalize the denominator". When the denominator is a cube root, you have to work harder to get it out of the bottom. Unfortunately, it is not as easy as choosing to multiply top and bottom by the radical, as we did in Example 2. Here is why:
mdv13
In the first case, the power of 2 and the index of 2 allow for a perfect square under a square root and the radical can be removed.
In the second case, the power of 2 with an index of 3 does not create an inverse situation and the radical is not removed. We need an additional factor of the cube root of 4 to create a power of 3 for the index of 3.
mdv14
It may be the case that the radicand of the cube root is simple enough to allow you to "see" two parts of a perfect cube hiding inside. In this case, you can simplify your work and multiply by only one additional cube root. As shown below, one additional factor of the cube root of 2, creates a perfect cube in the radicand.
mdv10



expin4 md3 ANSWER: md6
 

We will use a conjugate to rationalize the denominator! The problem with this fraction is that the denominator contains a radical. As such, the fraction is not considered to be in simplest form. As we saw in Example 8 above, multiplying a binomial times its conjugate will rationalize the product. This process will remove the radical from the denominator in this problem ( if we multiply the denominator by 1 + rad5). So as not to "change" the value of the fraction, we will multiply both the top and the bottom by 1 + rad5, thus multiplying by 1.

reduceradbottom
While the numerator "looks" worse, the denominator is now a rational number and the fraction is deemed in simplest form. Notice that there is nothing further we can do to simplify the numerator.

Did you notice how the process of "rationalizing the denominator" by using a conjugate resembles the "difference of squares": a2 - b2 = (a + b)(a - b)? They both create perfect squares, and eliminate any "middle" terms.

 

expin5 md7 ANSWER: ms9
md10

While the conjugate proved useful in the last problem when dealing with a square root in the denominator, it is not going to be helpful with a cube root in the denominator. Watch what happens when we multiply by a conjugate:
ms8
The cube root of 9 is not a perfect cube and cannot be removed from the denominator. Instead of removing the cube root from the denominator, the conjugate simply created a new cube root in the denominator.

To solve this problem, we need to think about the "sum of cubes formula":
a3 + b3 = (a + b)(a2 - ab + b2). This formula shows us that to obtain perfect cubes we need to multiply by more than just a conjugate term.
Let a = 1 and b = the cube root of 3. We will multiply top and bottom by
a2 - ab + b2.

md8

WOW!

Note: If the denominator had been 1 "minus" the cube root of 3, the "difference of cubes formula" would have been used: a3 - b3 = (a - b)(a2 + ab + b2).

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Expressions with Variables
expin1 mdv1 ANSWER:    6x2
  Multiplying will yield two perfect squares.
mdv2


expin2 mdv3 ANSWER: mdv17
  Search out the perfect cubes and reduce. This will simplify the multiplication. If you do not "see" the perfect cubes, multiply through and then reduce.
mdv16



expin3 mdv18 ANSWER: mdv20
  Using the approach we saw in Example 3 under Division, we multiply by two additional factors of the denominator. Look for perfect cubes in the radicand as you multiply to get the final result.
mdv19
Or, another approach is to create the simplest perfect cube under the radical in the denominator.
cube55

 

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