## anonymous 5 years ago Would the following fraction be considered to be fully simplified? x^4-2x^3-5x^2-18x-36 ----------------------------- x^5-5x^4+6x^3+8x^2-40x+48

1. anonymous

This was the result I got when multiplying two fractions. I know my multiplication was correct.

2. anonymous

No.

3. anonymous

Why?

4. anonymous

I'm guessing you'd have to solve the top and the bottom separately for it to be simplified.

5. anonymous

No, there is no solving. My instructions are to simplify completely. This is where I am in the problem. Does that mean I need to combine all of the x's and the whole numbers in front of them, and then subtract the smaller total from the larger total, and get an x with a single variable on whatever side of the equation that happens to be?

6. anonymous

Or just the -36/48 part...

7. anonymous

I believe its just the -36/48 part.

8. anonymous

Okay yea, so then that would make the final answer: $x^4-2x^3-5x^2-18x-3$ ------------------------ $x^5-5x^4+6x^3+8x^2-40x+12$

9. anonymous

?

10. anonymous

Lost me their, man. I'm only in 8th grade.

11. anonymous

Eh... Does that look completely simplified to you then?

12. anonymous

No. Doesn't look simplified. I would just try out all the options and see which one does look like its in its final "simplified" form...

13. anonymous

yo no entiendo las cosas que este chico esta preguntando me...

14. anonymous

Okay, then how about the original question... See what you can come up with... $(x^2-2x+4)/(x^2-5x+6)\div(x^2-9)/(x^3+8)$

15. anonymous

K.

16. anonymous

Darn it. I messed it up. I can't solve that right now. I'm studying at the same time I do this.

17. anonymous

Sorry man. You're on your own until someone with the proper knowledge can help you. :/

18. anonymous

Si no sabes cómo hacerlo, ¿por qué estás tratando de ayudar?

19. anonymous

Is this the original expression ${x^2-2x+4 \over x^2-5x+6}\div {x^2-9 \over x^3+8}?$

20. anonymous

yes

21. anonymous

At the very top is what I got from long multiplication. I just need to know if that form is correctly simplified since the exponents on the variables aren't all the same, or if I have to simplify those variables with the same exponents if possible.

22. anonymous

It does not look difficult at all. Just a minute.

23. anonymous

such as -2x^3/6x^3, would you simplify that to x^3/3x^3?

24. anonymous

I forgot the negative in that answer, sorry

25. anonymous

Are you sure it's division, not multiplication?

26. anonymous

Absolutely positive

27. anonymous

I flipped the fraction on the right side to turn the division sign into a multiplication sign.

28. anonymous

No problem. $={x^2-2x+4 \over (x-3)(x-2)}.{(x+2)(x^2-2x+4) \over (x-3)(x+3)}$ ${(x^2-2x+4)^2(x+2) \over (x-3)^2(x+3)(x-2)}$ For me, this is the best simplification.

29. anonymous

Ahhh, I got on here yesterday to ask about factoring it and someone told me not to do that.

30. anonymous

But you know this type of question usually involve some cancellation, which would be the case if it was a multiplication.

31. anonymous

But I get why you did it. It IS more simple in factored form.

32. anonymous

From what I remember of my first attempt at factoring it, all of the factored trinomials did not cancel.

33. anonymous

I have the work around here, just one secon.

34. anonymous

Yeah, nothing can be cancelled. Is this a question from your teacher or from the book?

35. anonymous

It is a study guide for my final exam for college algebra. There are no answers given, so I want to make sure I have the idea down.

36. anonymous

I see. Good luck. I'll be around of you have any questions that I have answers for.

37. anonymous

But in that form you gave, couldn't you factor it further?

38. anonymous

Nope.

39. anonymous

but x^2-2x+4 would be (x-2)(x-2) wouldn't it?

40. anonymous

Wait no, it won't work that way.

41. anonymous

But I get what you are saying. Factoring would be the simplest way to completely simplify a problem with two fractions and any kind of operation.

42. anonymous

Well there is one thing you could do to the denominator. The denominator can be written as (x^2-9)(x-3)(x-2)

43. anonymous

Actually, (x^2-2x+4)/1 would cross cancel with 1/(x+2)(x^2-2x+4) which is the result of x^3+8, right?

44. anonymous

But what I wrote before is more simplified than this, I would say.

45. anonymous

Yes, and I appreciate the help very much, I will let you move on to other things, but you helped me figure something out :D

46. anonymous

You're welcome, but you can't cancel them. :)

47. anonymous

if there was multiplication in the center you could though, right?

48. anonymous

Yeah.

49. anonymous

You could cancel many things actually.

50. anonymous

Like this: $(x^2-2x+4) * (1)$ $(1) (x+2)(x^2-2x+4)$

51. anonymous

eh, everything after the 1 is on the right side of that problem. Cant figure out how to like up the equation things

52. anonymous

It would seem like, in that scenario, you would have 1/1 * 1/(x+2)

53. anonymous

I don't get it :)

54. anonymous

Sorry, try it this way: 1 5 - * - 5 1

55. anonymous

Yeah this is 1.

56. anonymous

I am applying that principle to the fractions. You helped me figure it out though :D Thank you again!

57. anonymous

:)