anonymous
  • anonymous
Simplify the rational expression. State any restrictions on the variable.
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!
anonymous
  • anonymous
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anonymous
  • anonymous
@Nnesha do you know anything about this?
anonymous
  • anonymous
@jim_thompson5910

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jim_thompson5910
  • jim_thompson5910
what two numbers multiply to 24 and add to -10?
anonymous
  • anonymous
12 and 2
jim_thompson5910
  • jim_thompson5910
nope
jim_thompson5910
  • jim_thompson5910
12+2 = 14 not -10
anonymous
  • anonymous
|dw:1444458852967:dw| I believe this is the answer
anonymous
  • anonymous
opps for some reason I thought i read multiply lol thats what I get for 15 hours of school work today
jim_thompson5910
  • jim_thompson5910
-6 and -4 fit the description, no? -6 + (-4) = -10 -6 times -4 = 24
anonymous
  • anonymous
Yes that is correct. I see what you mean by that now
jim_thompson5910
  • jim_thompson5910
so that means n^4 - 10n^2 + 24 factors to (n^2 - 6)(n^2 - 4) we can then further factor n^2 - 4 down into (n-2)(n+2)
jim_thompson5910
  • jim_thompson5910
so overall, n^4 - 10n^2 + 24 completely factors to (n^2 - 6)(n-2)(n+2)
jim_thompson5910
  • jim_thompson5910
you'll do the same for the denominator and see if you can cancel anything
anonymous
  • anonymous
Okay so the bottom line reads n^4-9n^2+18 Im confused:/ Would I need to find a number that multiplies to 24 and adds to -9?
jim_thompson5910
  • jim_thompson5910
multiplies to 18 actually
jim_thompson5910
  • jim_thompson5910
find two numbers that multiply to 18 and add to -9
anonymous
  • anonymous
so 6 and 3 right cause 6*3=18 and -6+-3=-9
jim_thompson5910
  • jim_thompson5910
-6 and -3
jim_thompson5910
  • jim_thompson5910
I think that's what you meant to say?
anonymous
  • anonymous
right
jim_thompson5910
  • jim_thompson5910
so n^4-9n^2+18 factors to (n^2-6)(n^2-3)
jim_thompson5910
  • jim_thompson5910
n^4 - 10n^2 + 24 completely factors to (n^2 - 6)(n-2)(n+2) n^4-9n^2+18 completely factors to (n^2-6)(n^2-3)
jim_thompson5910
  • jim_thompson5910
at this point, I see a pair of terms cancelling
anonymous
  • anonymous
n^2-6
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
so (n^2-2) (n+2) (n^2-3) (n^2-4) (n^2-3)
jim_thompson5910
  • jim_thompson5910
I think you meant to say `(n-2) (n+2)` in your first step
jim_thompson5910
  • jim_thompson5910
but yeah the final answer is \[\Large \frac{n^2-4}{n^2-3}\]
jim_thompson5910
  • jim_thompson5910
what would the restrictions be?
anonymous
  • anonymous
so would the answer be this |dw:1444460035990:dw|
jim_thompson5910
  • jim_thompson5910
close
jim_thompson5910
  • jim_thompson5910
but no
anonymous
  • anonymous
could we work the problem out further? I quess i still dont 100% understand
jim_thompson5910
  • jim_thompson5910
the denominator factors into (n^2-6)(n^2-3) each piece is set equal to 0 and solved for to get n^2-6 = 0 n^2 = 6 n = sqrt(6) or n = -sqrt(6) n^2-3 = 0 n^2 = 3 n = sqrt(3) or n = -sqrt(3)
anonymous
  • anonymous
would it be the square root of those numbers ( 6, -3)
jim_thompson5910
  • jim_thompson5910
so the restrictions are actually \[\Large n \ne \pm \sqrt{6}\] \[\Large n \ne \pm \sqrt{3}\]
anonymous
  • anonymous
|dw:1444460209535:dw|
jim_thompson5910
  • jim_thompson5910
much better
anonymous
  • anonymous
thank you so much!! You are a lifesaver and such a great teacher! My husband usually helps me with math but he doesn't know this type of stuff! Im giving you a medal for your help.
jim_thompson5910
  • jim_thompson5910
I'm glad it's making more sense now

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