How to factor the expression completely:
((1+(4/x))^2)+((1-(4/x))^2)

- anonymous

How to factor the expression completely:
((1+(4/x))^2)+((1-(4/x))^2)

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- schrodinger

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- Mertsj

\[(1+(\frac{4}{x})^2)+(1-(\frac{4}{x})^2)\]

- Mertsj

Is that the problem?

- anonymous

Yes it is

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## More answers

- anonymous

Actually the (^2) is on the outside of the parentheses.

- Mertsj

\[(1+\frac{4}{x})^2+(1-\frac{4}{x})^2\]

- Mertsj

Is that it?

- anonymous

Yes! Sorry for the confusion..

- Mertsj

Expand each binomial and collect the like terms.

- anonymous

Does that simplify to 16/x?

- anonymous

I am so sorry again, but there's supposed to be a minus between the two quantities.. I got mixed up after all the parentheses

- Mertsj

\[1+\frac{8}{x}+\frac{16}{x^2}+1-\frac{8}{x}+\frac{16}{x^2}\]

- Mertsj

Collect the like terms.

- anonymous

With the minus in between I get 16/x? Is that the answer then?

- Mertsj

yes

- anonymous

Thank you so much!

- Mertsj

yw

- anonymous

Could you help with another problem?

- Mertsj

Only if you double check to make sure you have posted it correctly before I work on it.

- anonymous

I will. I'll attach the file, thank you.

- anonymous

The problem is in the attachment

##### 1 Attachment

- Mertsj

Factor out:
\[(\frac{1}{2}x ^{-\frac{1}{2}})(7x+4)^{-\frac{1}{2}}\]

- anonymous

That's the first portion of the problem right?

- Mertsj

It says to factor completely. That is what I would do as a first step is factor out those two common factors.

- anonymous

How do I factor out the (7x+4)^(-1/2) if the first one is positive?

- Mertsj

|dw:1377913408303:dw|

- Mertsj

What goes first inside the brackets that will result in the original first term when it is multiplied by the two factors outside the brackets?

- anonymous

Would that be (1)^(-1)?

- Mertsj

|dw:1377913939711:dw|

- Mertsj

Now do the same for the next term

- anonymous

When the quantities are multiplied, the exponents are added together?

- Mertsj

yes

- anonymous

So the next term in the brackets would be 7x^1

- Mertsj

|dw:1377914256550:dw|

- anonymous

Can it be further factored? Or no because the 7x+4 is in the parentheses?

- Mertsj

|dw:1377914423147:dw|

- Mertsj

Since 7x+4-7x=4

- Mertsj

|dw:1377914549187:dw|

- anonymous

Oh, okay. I see how it is now, and you did that because you can't find the factors because its an expression and not an equation, right?

- Mertsj

We did find the factors. You can't find what x equals because it is an expression and not an equation.

- anonymous

Okay. and if I was to write that in radical form, would it be \[\frac{2\sqrt{7x ^{2}+4x}}{7x^{2}+4x}\]

- Mertsj

yes

- anonymous

The program I using for practice, didn't recognize it as the right answer...

- Mertsj

Didn't recognize what? The last thing you posted...simplified radical form?

- anonymous

Yeah, it said it was incorrect

- anonymous

And it didn't recognize it with the exponents either

- Mertsj

Did you try
\[\frac{2}{\sqrt{x(7x+4)}}\]

- anonymous

Yes I did

- Mertsj

How about:
\[2x ^{-\frac{1}{2}}(7x+4)^{-\frac{1}{2}}\]

- anonymous

Yeah I tried with the exponents

- Mertsj

Or
\[2(x(7x+4))^{-\frac{1}{2}}\]

- Mertsj

Or
\[2(7x^2+4x)^{-\frac{1}{2}}\]

- anonymous

I have to put the (^1/2) as radical when trying to submit the answer

- Mertsj

http://www.wolframalpha.com/input/?i=Factor%3A++.5x^%28-.5%29%287x%2B4%29^.5-3.5x^.5%287x%2B4%29^-.5

- Mertsj

\[\frac{2}{\sqrt{x}\sqrt{7x+4}}\]

- Mertsj

Try that.

- anonymous

That worked! Is the one with no radical in the denominator different somehow or is the program just picky with the answer?

- anonymous

Thank you for all your help too.

- Mertsj

The program is just picky with the answer. If you would read your entry directions carefully, it might give you some indication of the way to enter them. I have even had the order of the factors be an issue from time to time.

- Mertsj

That's why I went to Wolfram. Sometimes these computerized programs use computers to solve the problems so I thought the answers might be in the same form.

- anonymous

Okay that's good know, thank you again!

- Mertsj

yw

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