## anonymous one year ago Simplify the rational expression. State any restrictions on the variable.

1. anonymous

2. anonymous

3. anonymous

@jim_thompson5910

4. jim_thompson5910

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

5. anonymous

12 and 2

6. jim_thompson5910

nope

7. jim_thompson5910

12+2 = 14 not -10

8. anonymous

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

9. anonymous

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

10. jim_thompson5910

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

11. anonymous

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

12. 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)

13. jim_thompson5910

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

14. jim_thompson5910

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

15. 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?

16. jim_thompson5910

multiplies to 18 actually

17. jim_thompson5910

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

18. anonymous

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

19. jim_thompson5910

-6 and -3

20. jim_thompson5910

I think that's what you meant to say?

21. anonymous

right

22. jim_thompson5910

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

23. 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)

24. jim_thompson5910

at this point, I see a pair of terms cancelling

25. anonymous

n^2-6

26. jim_thompson5910

yes

27. anonymous

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

28. jim_thompson5910

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

29. jim_thompson5910

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

30. jim_thompson5910

what would the restrictions be?

31. anonymous

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

32. jim_thompson5910

close

33. jim_thompson5910

but no

34. anonymous

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

35. 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)

36. anonymous

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

37. jim_thompson5910

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

38. anonymous

|dw:1444460209535:dw|

39. jim_thompson5910

much better

40. 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.

41. jim_thompson5910

I'm glad it's making more sense now