anonymous
  • anonymous
How to factor the expression completely: ((1+(4/x))^2)+((1-(4/x))^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mertsj
  • Mertsj
\[(1+(\frac{4}{x})^2)+(1-(\frac{4}{x})^2)\]
Mertsj
  • Mertsj
Is that the problem?
anonymous
  • anonymous
Yes it is

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anonymous
  • anonymous
Actually the (^2) is on the outside of the parentheses.
Mertsj
  • Mertsj
\[(1+\frac{4}{x})^2+(1-\frac{4}{x})^2\]
Mertsj
  • Mertsj
Is that it?
anonymous
  • anonymous
Yes! Sorry for the confusion..
Mertsj
  • Mertsj
Expand each binomial and collect the like terms.
anonymous
  • anonymous
Does that simplify to 16/x?
anonymous
  • 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
  • Mertsj
\[1+\frac{8}{x}+\frac{16}{x^2}+1-\frac{8}{x}+\frac{16}{x^2}\]
Mertsj
  • Mertsj
Collect the like terms.
anonymous
  • anonymous
With the minus in between I get 16/x? Is that the answer then?
Mertsj
  • Mertsj
yes
anonymous
  • anonymous
Thank you so much!
Mertsj
  • Mertsj
yw
anonymous
  • anonymous
Could you help with another problem?
Mertsj
  • Mertsj
Only if you double check to make sure you have posted it correctly before I work on it.
anonymous
  • anonymous
I will. I'll attach the file, thank you.
anonymous
  • anonymous
The problem is in the attachment
1 Attachment
Mertsj
  • Mertsj
Factor out: \[(\frac{1}{2}x ^{-\frac{1}{2}})(7x+4)^{-\frac{1}{2}}\]
anonymous
  • anonymous
That's the first portion of the problem right?
Mertsj
  • Mertsj
It says to factor completely. That is what I would do as a first step is factor out those two common factors.
anonymous
  • anonymous
How do I factor out the (7x+4)^(-1/2) if the first one is positive?
Mertsj
  • Mertsj
|dw:1377913408303:dw|
Mertsj
  • 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
  • anonymous
Would that be (1)^(-1)?
Mertsj
  • Mertsj
|dw:1377913939711:dw|
Mertsj
  • Mertsj
Now do the same for the next term
anonymous
  • anonymous
When the quantities are multiplied, the exponents are added together?
Mertsj
  • Mertsj
yes
anonymous
  • anonymous
So the next term in the brackets would be 7x^1
Mertsj
  • Mertsj
|dw:1377914256550:dw|
anonymous
  • anonymous
Can it be further factored? Or no because the 7x+4 is in the parentheses?
Mertsj
  • Mertsj
|dw:1377914423147:dw|
Mertsj
  • Mertsj
Since 7x+4-7x=4
Mertsj
  • Mertsj
|dw:1377914549187:dw|
anonymous
  • 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
  • Mertsj
We did find the factors. You can't find what x equals because it is an expression and not an equation.
anonymous
  • 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
  • Mertsj
yes
anonymous
  • anonymous
The program I using for practice, didn't recognize it as the right answer...
Mertsj
  • Mertsj
Didn't recognize what? The last thing you posted...simplified radical form?
anonymous
  • anonymous
Yeah, it said it was incorrect
anonymous
  • anonymous
And it didn't recognize it with the exponents either
Mertsj
  • Mertsj
Did you try \[\frac{2}{\sqrt{x(7x+4)}}\]
anonymous
  • anonymous
Yes I did
Mertsj
  • Mertsj
How about: \[2x ^{-\frac{1}{2}}(7x+4)^{-\frac{1}{2}}\]
anonymous
  • anonymous
Yeah I tried with the exponents
Mertsj
  • Mertsj
Or \[2(x(7x+4))^{-\frac{1}{2}}\]
Mertsj
  • Mertsj
Or \[2(7x^2+4x)^{-\frac{1}{2}}\]
anonymous
  • anonymous
I have to put the (^1/2) as radical when trying to submit the answer
Mertsj
  • 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
  • Mertsj
\[\frac{2}{\sqrt{x}\sqrt{7x+4}}\]
Mertsj
  • Mertsj
Try that.
anonymous
  • anonymous
That worked! Is the one with no radical in the denominator different somehow or is the program just picky with the answer?
anonymous
  • anonymous
Thank you for all your help too.
Mertsj
  • 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
  • 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
  • anonymous
Okay that's good know, thank you again!
Mertsj
  • Mertsj
yw

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