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Simplify Radical Expressions using Prime Factorization

Everyone likes things that are easy to understand right? Just because something is easy to understand does not mean it is the quickest or simplest way. However we all need to start somewhere and I think starting with the basics using what you already know is a great start. When simplifying radical expressions by prime factorization we need to learn how to rewrite a number or variable exponent as a product of it’s prime factors. I just threw out a lot of vocabulary at you so what does it all mean?

Prime Number – A number that can only be divided by itself and one

ex) 5 ex) 11 ex) 29

Composite Number – A number that can be divided by another number other than itself and one

ex) 4 ex) 12 ex) 32

Factoring(a number) – Rewriting a number or as a product of two or more numbers

ex) $ 6=3times 2$

Prime Factorization – is rewriting a number or variable as a product of its prime factors

ex) $18=2times 3times 3$

ex) ${{x}^{5}}=x*x*x*x*x$

Factoring a number and the Prime Factorization of a number can be the same but not necessary always so

ex) Take the number 12. We can factor 12 as $6times 2$ but the prime factorization of 12 can only be written as, $2times 2times 3$

The first way we learned how to find all of the prime factors of a number was to create a factor tree. You can learn how to factor a number by using a factor tree here. Once we have the prime factorization of a number we can now simplify radical expressions but we need to know everything that makes up a radical expression first.

$sqrt[a]{b}$

Index/root(a) – the value that a number has to be multiplied by itself to equal the radicand

Radical symbol($sqrt{{}}$) – The symbol that expresses the operation 

Radicand(b) – The value that we are taking the root of

To put it simply taking a root is like asking your self “What value multiplied by itself as many times as the root is equal to the radicand?”

Does that make sense? What about if we apply it to a specific problem

ex) $sqrt[2]{25}$ is asking what number multiplied twice or by itself will be 25? Well the answer to that is 5 because 5times 5=25 which is our radicand

ex) $sqrt[3]{{{x}^{3}}}$ is asking what number multiplied three times by itself will be ${{x}^{3}}$? Well the answer to that is x because $xcdot xcdot x={{x}^{3}}$ which is our radicand

Where did this question come from? Did someone just make it up?

Well actually taking the root is the inverse operation of the power operation.
Just like adding/subtracting are inverse operations taking the root of a number and raising it to the same power are inverse operations. They undo each other. Here is a quick breakdown of what I am talking about.

Adding/Subtracting
If we start with 5 and add 3 we obtain 8. If we subtract 3 from 8 we end back up with our original number 5
$5to 5+3=8to 8-3=5$

Multiply/Dividing
If we start with 5 and multiply by 3 we obtain 15. If we divide 15 by 3 we end back up with our original number 5
$5to 5times 3=15to 15div 3=5$

Raising to a power/Taking the root
If we start with 5 and raise it to the power of 2 we obtain 25. If we take the root 2 (square root) of 25 we end back up with our original number 5
$5to {{5}^{2}}=25to sqrt[5]{25}=5$

You can see with all of these inverse operations that by applying both operations we end with the same value we started with. This is because inverse operations undo each other.When you are just taking the root of a number, you will want to first notice value of the index. If there is not a number there it is implied that the number is 2.

ex) $sqrt[2]{radicand}=sqrt{radicand}$

The reason I say look at the index first is because as the index changes so will your answer even if you have the same radicand!

ex) sqrt[2]{64}=sqrt[2]{8times 8}}=8

ex) sqrt[3]{64}=sqrt[3]{4times 4times 4}}=4

We can also take the root of variables

ex) $sqrt[4]{{{x}^{4}}}=sqrt[4]{xtimes xtimes xtimes x}}=x$

I want to make sure that I point out that to simplify the radical expressions I write the radicand as its prime factorization. This allows me to easily see how many times the factors are multiplied together. When I have them multiplied by each other the same number of times as the index I strike through them to signify I was able to take the root of the factored expression.What I have presented so far has been problems where the number of times the factors were multiplied by each other was the same as the value of the index. However this will not always be the case.

When we simplify a radical with extra factors we have to leave the leftover factors under the radicand.

ex) $sqrt[2]{8}= sqrt[2]{cancel{2times 2}times 2}= 2sqrt[2]{2}$

Again the square root asks us what number is multiplied by itself. Well in this example we only have one grouping of two’s multiplied together with an extra two. We take the square root of 2 times 2 (which is 2) and leave the other two under the radicand.
Sometimes we will be able to take the root of multiple sets factors. When that is the case simply multiply the simplified expressions together outside the radical.

ex) $sqrt[2]{243}=sqrt[2]{cancel{3times 3times 3times 3}times 3}=3times 3times sqrt[2]{3}=9sqrt[2]{3}$

In this example we are able to take the square root of two sets of 3’s multiplied by each other.
Now I have been pretty nice to you if you want to think of it that way. I have been using numbers that the prime factorization was always the same factors. However that is not always the case

ex) $sqrt[2]{50}=sqrt[2]{cancel{5times 5}times 2}=5sqrt[2]{2}$

ex) $sqrt[3]{72}=sqrt[3]{3times 3times cancel{2times 2times 2}}=2sqrt[3]{9}$

The same rule applies for variables even bigger expressions

ex) $sqrt[3]{{{x}^{11}}}=sqrt[3]{cancel{xtimes xtimes x}times cancel{xtimes xtimes x}times cancel{xtimes xtimes x}times xtimes x}=xtimes xtimes xtimes sqrt[3]{xtimes x}={{x}^{3}}sqrt[3]{{{x}^{2}}}$

And finally bringing numbers and variables together

ex) $sqrt[2]{12{{x}^{4}}{{y}^{3}}}=sqrt[2]{cancel{2times 2}times 3times cancel{xtimes x}times cancel{xtimes x}times cancel{ytimes y}times y}=2times xtimes xtimes ytimes sqrt[2]{3times y}=2{{x}^{2}}ysqrt[2]{3y}$

We could keep going but as you noticed once we start dealing with larger numbers and variables the prime factorization becomes very large. You can still always use this process if time is on your side. If you would like to know the some shortcuts to this process check out my other posts.