## Saylilbaby one year ago help pls im begging you all

1. anonymous

with?

2. pooja195

It helps i you actually asked your question.

3. pooja195

*if

4. saylilbaby

5. saylilbaby

6. saylilbaby

7. saylilbaby

@Michele_Laino

8. saylilbaby

@satellite73 @whpalmer4 @Zale101 @thomaster

9. whpalmer4

@saylilbaby are you here now?

10. saylilbaby

yes @whpalmer4

11. saylilbaby

@texaschic101

12. whpalmer4

Okay, let's look at the first one. Are you able to factor the numerator and denominator?

13. whpalmer4

$\frac{n^4-11n^2+30}{n^4-7n^2+10}$

14. whpalmer4

looking at the numerator, $$n^4-11n^2+30$$ what are two numbers which when multiplied give you $$30$$, but when added give you $$-11$$? Hopefully it is obvious that both must be negative numbers.

15. saylilbaby

is it C @whpalmer4 ?

16. saylilbaby

or A im confused @whpalmer4

17. whpalmer4

Let's work through the problem together, and you should be certain of your answer by the time we are done.

18. saylilbaby

i already worked thru i just needto state the restriction @whpalmer4

19. whpalmer4

Ok, didn't want to assume that...but it makes life easier :-) So, let's look at the factored denominator of the original expression: $n^4-7n^2+10 = (n^2-5)(n^2-2)$Agreed?

20. whpalmer4

When we find the restrictions, we have to do so on the unsimplified version, because we still have a "hole" in the function at those spots where we simplified away a term in the denominator. That means our restrictions on $$n$$ here are located at all values of $$n$$ where $n^2-5=0$and$n^2-2=0$ Any value of $$n$$ that makes either of those true will cause a division by $$0$$ in the original expression.

21. whpalmer4

We can see that A is not the right answer, because the restrictions it imposes are $$n\ne5$$ and $$n\ne2$$, but those values do not cause the original denominator to be equal to $$0$$. $(5^2-5)(5^2-2) = 460$$(2^2-5)(2^2-2) = -2$

22. whpalmer4

D is also not a correct answer for the same reason. $(5^2-5)(5^2-2) = 460$$(-2)^2-5)((-2)^2-2) = -2$

23. whpalmer4

B has the wrong fraction, so our hopes are left with C. Let's check it! $n\ne \pm \sqrt{5}$$n\ne\pm\sqrt{2}$ If we evaluate the original denominator at $$n=\sqrt{5}$$ $(\sqrt{5}^2-5)(\sqrt{5}^2-2) = (5-5)(5-2)= 0$ If we evaluate the original denominator at $$n=-\sqrt{5}$$ $((-\sqrt{5})^2-5)((-\sqrt{5})^2-2) = (5-5)(5-2)=0$ So far, so good. Now trying $$n=\pm\sqrt{2}$$ $(\sqrt{2}^2-5)(\sqrt{2}^2-2) = (2-5)(2-2) = 0$ $((-\sqrt{2})^2-5)((-\sqrt{2})^-2) = (2-5)(2-2) = 0$ Those are all good, so C is our answer.

24. whpalmer4

To find the restrictions, factor the initial denominator. Set each product term $$=0$$ and find all the values which satisfy that. The list of all of such values for all of the terms is the set of restrictions on the expression.