Simplify{[x^(5/3)]-[x^(1/3)]*[z^(4/3)]}/{[x^(2/3)]+[z^(2/3)]

[cjcapta]

This problem is different than the others, because this is the only problem that has 2 differenet variables being added in the denominator. I need a walkthrough on this problem:

The directions say: Simplify each expression.
{[x^(5/3)] - [x^(1/3)] * [z^(4/3)]} / {[x^(2/3)] + [z^(2/3)]}

 

[mmm4444bot]

Properties of Exponents and Factorization Patterns ...

... this is the only problem that has 2 differenet variables being added in the denominator ...

Hello CJ:

Two variables in a given expression are common in many algebra exercises. If you're not comfortable with this, then now is the time to start becoming so!

Consider the following expression.

\(\displaystyle 3x \;+\; 3x \cdot z\)

This expression contains two variables. Can you factor it to get the following?

\(\displaystyle 3x \cdot (1 \;+\; z)\)

How about the following expression which also contains two variables?

\(\displaystyle x^5 \;-\; x \cdot z^4\)

It also factors.

\(\displaystyle x \cdot (x^4 \;-\; z^4)\)

If you understand these examples, then perhaps you are not as befuddled by multiple variables as you think you are.

The exercise that you posted is not much different; the exponents are fractional, but this should not be a big concern since they are all THIRDS. In other words, instead of adding whole numbers in the exponent position, you'll be adding some number of thirds in the exponent position as you think through the factorizations.

EXAMPLE:

\(\displaystyle x^{1/3} \cdot x^{4/3} \;=\; x^{1/3 + 4/3} \;=\; x^{5/3}\)

VERSUS:

\(\displaystyle x^1 \cdot x^4 \;=\; x^{1 + 4} \;=\; x^5\)

So, we see that fractional exponents are not too difficult to handle.

I read your exercise as the following.

\(\displaystyle \frac{x^{5/3} \;-\; x^{1/3} \cdot z^{4/3}}{x^{2/3} + z^{2/3}}\)

We need three skills to simplify this one.

1) Factoring with fractional exponents
2) The property of exponents that deals with a power of a power
3) The factorization pattern for a Difference of Squares

I'll make up a similar exercise in which the steps are the same. Instead of cube roots (i.e., thirds), I'll use seventh roots (i.e., sevenths). I'll also change the variable names just to mix it up a bit more. Other than these changes, my example will use the same steps to simplify as I would use on your posted exercise.

\(\displaystyle \frac{a^{5/7} \;-\; a^{1/7} \cdot b^{4/7}}{a^{2/7} + b^{2/7}}\)

1) Note in the numerator that both of the terms a[sup:1cnu0pvt]5/7[/sup:1cnu0pvt] and a[sup:1cnu0pvt]1/7[/sup:1cnu0pvt] contain a factor of a[sup:1cnu0pvt]1/7[/sup:1cnu0pvt]. Factor out this term.

\(\displaystyle \frac{a^{1/7} \cdot (a^{4/7} \;-\; b^{4/7})}{a^{2/7} + b^{2/7}}\)

2) Recognize that the expression inside the parentheses is a Difference of Squares. This becomes apparent when you realize that raising any base to the 4/7ths power is the same as first raising it to the 2/7ths power, followed by squaring the result.

EXAMPLE:

\(\displaystyle a^{4/7} = (a^{2/7})^2\)

This comes from the Property of Exponents that deals with raising a power to another power.

x[sup:1cnu0pvt](n*m)[/sup:1cnu0pvt] = (x[sup:1cnu0pvt]n[/sup:1cnu0pvt])[sup:1cnu0pvt]m[/sup:1cnu0pvt]

And, of course, 4/7 = (2/7)*(2)

Rewrite the fourth powers as squares being squared.

\(\displaystyle \frac{a^{1/7} \cdot ([a^{2/7}]^2 \;-\; [b^{2/7}]^2)}{a^{2/7} + b^{2/7}}\)

3) Now the expression inside the parentheses is clearly seen as a Difference of Squares. Factor it using the factorization pattern for a Difference of Squares.

\(\displaystyle \frac{a^{1/7} \cdot (a^{2/7} \;+\; b^{2/7}) \cdot (a^{2/7} \;-\; b^{2/7})}{a^{2/7} + b^{2/7}}\)

The reason we factored the Difference of Squares to begin with is that we first recognized the denominator as one of the two factors in this special factorization pattern! That's nice because it means that we have a way to cancel the denominator (i.e., it's a great way to simplify).

\(\displaystyle a^{1/7} \cdot (a^{2/7} \;-\; b^{2/7})\)

This last result above should be acceptable, but, if we prefer, we can use the Distributive Property to execute the multiplication.

\(\displaystyle a^{3/7} \;-\; a^{1/7} \cdot b^{2/7}\)

Well, I practically did your exercise for you, but you should pay close attention to my steps anyway; firstly, because that's how you learn, and, secondly, because I make lots of mistakes. If you need more help simplifying your exercise, then please show whatever work you've been able to accomplish, and try to say something about why you're stuck.

Cheers,

~ Mark

 

[cjcapta]

Re: Need help with simplifying radical exponent expressions.

I have a couple of questions about your problem, but first I need to know how you can type your math problem so it actually looks like the problem?

 

[mmm4444bot]

LaTex ...

... I need to know how you can type your math problem so it actually looks like the problem

Hey there:

Are you asking how to get this web site to display mathematical notation (eg: symbols, fractions, exponents, etc.)?

If so, then you are asking how to type LaTex coding.

You can learn about this in two ways.

1) If you click the

button on any post that contains LaTex expressions, then you can see the actual code that the poster used.

2) If you click on the drop-down menu [Forum Help] at the top of a page, and select the menu item [Getting Started w/ LaTex], then you'll be taken to a page with a bunch of links for various examples. (Be forewarned! There are different variations of LaTex coding provided at these links; some of these work at this site, and some of do not.)

Here's the two main LaTex codes that I used in my post.

x^{1/3} displays fractional exponents.

\frac{numerator}{denominator} formats a fraction.

You can nest LaTex codes within each other.

\frac{x^{2/3} + z^{4/3}}{x^{1/3} - z^{2/3}} yields the following.

\(\displaystyle \frac{x^{2/3} + z^{4/3}}{x^{1/3} - z^{2/3}}\)

Of course, each line of LaTex coding must begin with [ tex ] and end with [ /tex ] -- do not type spaces around the square brackets; I typed the spaces here to prevent the web site from interpreting this example as actual LaTex code.

ALWAYS CLICK THE PREVIEW BUTTON TO SEE WHAT YOUR CODING WILL LOOK LIKE BEFORE SUBMITTING YOUR POST. (It's very easy to create a LaTexMess!)

Cheers,

~ Mark

PS: The expression that you typed in your original post is also just fine; you do not actually need to use LaTex as long as you're as careful in showing the grouping symbols as you have been so far.

MY EDIT: minor grammatical correction

 

[Denis]

Re: Need help with simplifying radical exponent expressions.

{[x^(5/3)] - [x^(1/3)] * [z^(4/3)]} / {[x^(2/3)] + [z^(2/3)]}

Sometimes easier to simplify the problem itself...just a suggestion:

let a = 1/6 ; then your expression becomes:
[x^(10a) - x^(2a) z^(8a)] / [x^(4a)] + [z^(4a)] ; since a^2 - b^2 = (a + b)(a - b) as per Mark's suggestion:

= [x^(5a) + x^(a) z^(4a)] [x^(5a) - x^(a) z^(4a)] / [x^(4a)] + [z^(4a)]

= [x^(a)][x^(4a) + x^(0) z^(4a)] [x^(5a) - x^(a) z^(4a)] / [x^(4a)] + [z^(4a)] ; remember that x^0 = 1 :

= [x^(a)] [x^(5a) - x^(a) z^(4a)]

= x^(6a) - x^(2a)z^(4a) ; substitute back in:

= x - x^(1/3)z^(2/3)

I'm too lazy to learn LaTex :shock:

 

[cjcapta]

Re: Properties of Exponents and Factorization Patterns ...

You said :

\(\displaystyle \frac{a^{5/7} \;-\; a^{1/7} \cdot b^{4/7}}{a^{2/7} + b^{2/7}}\)

1) Note in the numerator that both of the terms a[sup:17evilar]5/7[/sup:17evilar] and a[sup:17evilar]1/7[/sup:17evilar] contain a factor of a[sup:17evilar]1/7[/sup:17evilar]. Factor out this term.

\(\displaystyle \frac{a^{1/7} \cdot (a^{4/7} \;-\; b^{4/7})}{a^{2/7} + b^{2/7}}\)

But it doesn't look like you factored out an a[sup:17evilar]1/7[/sup:17evilar]

Shouldn't it be \(\displaystyle \frac{a^{1/7} \cdot (1^{4/7} \;-\; 1)\cdot (b^{4/7})}{a^{2/7} + b^{2/7}}\) ?

 

[Denis]

Re: Need help with simplifying radical exponent expressions.

NO, it shouldn't; Mark is correct.

Take a simpler case: a^5 - a^3 * b^4

That factors to: a^3(a^2 - b^4)

a^3 * a^2 = a^(3 + 2) = a^5

 

[cjcapta]

Re: Need help with simplifying radical exponent expressions.

NO, it shouldn't; Mark is correct.

Take a simpler case: a^5 - a^3 * b^4

That factors to: a^3(a^2 - b^4)

a^3 * a^2 = a^(3 + 2) = a^5

Okay I get that, and I get what Mark was saying now, but won't you get the same answer if you do it the way I said? :?:

 

[Denis]

Re: Need help with simplifying radical exponent expressions.

NO.
1^(4/7) = 1, so you'll get a result of 0.

 

[cjcapta]

Re: Properties of Exponents and Factorization Patterns ...

I meant to use the distributive property, wasn't the goal of what Mark is saying is to find a gcf? \(\displaystyle \frac{a^{1/7} \cdot (1^{4/7}) \;-\; (a^{1/7})(1)\cdot (b^{4/7})}{a^{2/7} + b^{2/7}}\)

Although, I think I understand now, but I think that if it wasn't 1 then you could do it. Don't you add the exponents when multiplying or do the bases need to be the same for multiplying?

 

[mmm4444bot]

Re: Properties of Exponents and Factorization Patterns ...

I meant to use the distributive property, wasn't the goal of what Mark is saying is to find a gcf? \(\displaystyle \frac{a^{1/7} \cdot (1^{4/7}) \;-\; (a^{1/7})(1)\cdot (b^{4/7})}{a^{2/7} + b^{2/7}}\)

Although, I think I understand now, but I think that if it wasn't 1 then you could do it. Don't you add the exponents when multiplying or do the bases need to be the same for multiplying?

Hello CJ:

You changed the first term in my example from a[sup:1qxxz35s]5/7[/sup:1qxxz35s] to a[sup:1qxxz35s]1/7[/sup:1qxxz35s], and then you multiplied it by 1[sup:1qxxz35s]4/7[/sup:1qxxz35s].

I'm not sure what you're trying to do by making these changes.

The Greatest Common Factor in the original numerator of my example is a[sup:1qxxz35s]1/7[/sup:1qxxz35s]. It's also the ONLY common factor; there is no other factorization that can be done in the numerator.

I'll retype my example without the seventh roots; perhaps you will gain something from seeing this new example.

\(\displaystyle \frac{a^5 - a \cdot b^4}{a^2 + b^2}\)

\(\displaystyle \frac{a \cdot (a^4 - b^4)}{a^2 + b^2}\)

\(\displaystyle \frac{a \cdot (a^2 + b^2) \cdot (a^2 - b^2)}{a^2 + b^2}\)

\(\displaystyle a \cdot (a^2 - b^2)\)

The answers to your two latest questions are 'yes' and 'yes'.

Yes, when you multiply two terms with exponents together, you add the exponents.

Yes, the bases must be the same before you can multiply terms with exponents using this exponent rule.

Cheers,

~ Mark

 

[cjcapta]

Re: Properties of Exponents and Factorization Patterns ...

Yes, the bases must be the same before you can multiply terms with exponents using this exponent rule.

Cheers,

~ Mark

So you can't multiply them? \(\displaystyle {a^{1/7} \cdot (1^{4/7}) \;-\; (a^{1/7})(1)}\) I think that was what I didn't understand for that part. I get that part now.

 

[mmm4444bot]

Re: Properties of Exponents and Factorization Patterns ...

Yes, the bases must be the same before you can multiply terms with exponents using this exponent rule.

Cheers,

~ Mark

So you can't multiply them? \(\displaystyle {a^{1/7} \cdot (1^{4/7}) \;-\; (a^{1/7})(1)}\) I think that was what I didn't understand for that part. I get that part now.

Hi CJ:

When you ask about "them", you are using an unreferenced pronoun. Your question is therefore ambiguous, but I will try to guess over what you're trying to ask about, nonetheless.

I stated the following.

... Yes, the bases must be the same before you can multiply terms with exponents using this exponent rule.

Please note my added emphasized on the phrase "using this exponent rule".

This rule regards adding exponents when multiplying two exponential terms. You may only add the exponents when the bases are the same.

Therefore, NO. We may not use this rule when multiplying a[sup:2kuu3utd]1/7[/sup:2kuu3utd] times 1[sup:2kuu3utd]4/7[/sup:2kuu3utd] to get a[sup:2kuu3utd]5/7[/sup:2kuu3utd] because the bases are not the same.

Of course, we may always multiply anything by 1[sup:2kuu3utd]4/7[/sup:2kuu3utd], but I really have no idea what you are trying to accomplish by writing 1[sup:2kuu3utd]4/7[/sup:2kuu3utd] in the first place.

Were you able to simply your original exercise?

Cheers,

~ Mark