Simplify the rational expression. State any restrictions on the variable.

- anonymous

Simplify the rational expression. State any restrictions on the variable.

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- anonymous

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- anonymous

@Nnesha do you know anything about this?

- anonymous

@jim_thompson5910

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## More answers

- jim_thompson5910

what two numbers multiply to 24 and add to -10?

- anonymous

12 and 2

- jim_thompson5910

nope

- jim_thompson5910

12+2 = 14 not -10

- anonymous

|dw:1444458852967:dw| I believe this is the answer

- anonymous

opps for some reason I thought i read multiply lol thats what I get for 15 hours of school work today

- jim_thompson5910

-6 and -4 fit the description, no?
-6 + (-4) = -10
-6 times -4 = 24

- anonymous

Yes that is correct. I see what you mean by that now

- 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

so overall, n^4 - 10n^2 + 24 completely factors to (n^2 - 6)(n-2)(n+2)

- jim_thompson5910

you'll do the same for the denominator and see if you can cancel anything

- 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

multiplies to 18 actually

- jim_thompson5910

find two numbers that multiply to 18 and add to -9

- anonymous

so 6 and 3 right cause 6*3=18 and -6+-3=-9

- jim_thompson5910

-6 and -3

- jim_thompson5910

I think that's what you meant to say?

- anonymous

right

- jim_thompson5910

so n^4-9n^2+18 factors to (n^2-6)(n^2-3)

- 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

at this point, I see a pair of terms cancelling

- anonymous

n^2-6

- jim_thompson5910

yes

- anonymous

so (n^2-2) (n+2)
(n^2-3)
(n^2-4)
(n^2-3)

- jim_thompson5910

I think you meant to say `(n-2) (n+2)` in your first step

- jim_thompson5910

but yeah the final answer is \[\Large \frac{n^2-4}{n^2-3}\]

- jim_thompson5910

what would the restrictions be?

- anonymous

so would the answer be this |dw:1444460035990:dw|

- jim_thompson5910

close

- jim_thompson5910

but no

- anonymous

could we work the problem out further? I quess i still dont 100% understand

- 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

would it be the square root of those numbers ( 6, -3)

- jim_thompson5910

so the restrictions are actually
\[\Large n \ne \pm \sqrt{6}\]
\[\Large n \ne \pm \sqrt{3}\]

- anonymous

|dw:1444460209535:dw|

- jim_thompson5910

much better

- 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

I'm glad it's making more sense now

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