## algebra2sucks 2 years ago What are the points of discontinuity? Are they all removable? y = (x-3)/x^2 - 12x + 27

1. myininaya

When is the bottom 0?

2. myininaya

For what values of x make the bottom 0?

3. algebra2sucks

not sure?

4. myininaya

So you should be able to solve an equation in this form ax^2+bx+c=0 by now. I'm guessing this since this is a type of problem that occurs after learning that.

5. myininaya

Do you know how to factor?

6. myininaya

7. myininaya

You can even do completing the square.

8. algebra2sucks

so x^2+12x+27=0?

9. algebra2sucks

how do i do that?

10. myininaya

Do you have -12x or +12x?

11. algebra2sucks

its postive 12x

12. myininaya

Hmmm...So the problem above is written/typed wrongly I assume. Okay You should know how to factor the polynomial x^2+12x+27

13. myininaya

Find two factors of c that add up to be b (you can do this when the coefficient of x^2 is 1 which in this problem it is)

14. myininaya

In other words find two factors of 27 that add up to be 12.

15. algebra2sucks

its like this |dw:1370027288868:dw|

16. myininaya

err so it is -12x?

17. algebra2sucks

noo x^2 minus 12x

18. myininaya

I'm really confused what problem we are doing because twice now you have written +12x

19. myininaya

You have also said twice that we have -12x

20. myininaya

Anyways, we take the bottom polynomial whatever it may be and set that equal to zero and solve for x. I will let you do it since I don't know what problem we are doing anymore.

21. algebra2sucks

sorry

22. algebra2sucks

this is the problem were doing

23. myininaya

Set bottom polynomial equal to zero. $x^2-12x+27=0$ You will notice this is in the form $ax^2+bx+c=0$ You have learned if you cannot factor this easily you could also use completing the square or quadratic formula. The easy way to solve this quadratic is by factoring in my opinion. Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n. So The factored form would be (x-m)(x-n)=0 Then you set both factors equal to 0 like so x-m=0 or x-n=0 Solve both for x like so x=m or x=n

24. algebra2sucks

so which numbers do we use for x-m and x-n?

25. myininaya

Have you found factors of 27 that have product 27 and sum -12?

26. myininaya

Those are your m and n.

27. myininaya

Recall this part that I said: Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n.

28. myininaya

c is 27 and b is -12

29. myininaya

|dw:1370027980807:dw|

30. myininaya

See how I factored 27?

31. myininaya

1(27)=27 3(9)=27 -1(-27)=27 -3(-9)=27

32. algebra2sucks

Yes I see on my answes I have x= -9, x =-3 is that right?

33. myininaya

Do any of those factors above that multiplied to be 27 also add up to be -12?

34. algebra2sucks

are these numbers removable?

35. algebra2sucks

and -3?

36. myininaya

(x-3)(x-9)=0 Set both equal to 0.

37. myininaya

I put (x-m)(x-n)=0 where I meant (x+m)(x+n)=0. So you would have x+m=0 or x+n=0 Which means you would have x=-m or x=-n. But anyways we have (x-3)(x-9)=0 And you are to set both factors equal to 0 and solve for x for both equations

38. myininaya

x-3=0 or x-9=0 I leaving solving these linear equations to you.

39. algebra2sucks

do we also use -1 or no?

40. myininaya

Where does that come from?

41. myininaya

Have you solve x-3=0 or x-9=0 yet?

42. algebra2sucks

no idk how

43. myininaya

x-3=0 You can't solve this for x? You don't know how to get x by itself?

44. myininaya

To undo subtraction by three you add three.

45. myininaya

Try adding three to both sides and tell me what the result is.

46. myininaya

x-3=0 add 3 to both sides +3 +3 --------- what is the result?

47. myininaya

x-3+3=0+3

48. myininaya

what is -3+3? what is 0+3?

49. algebra2sucks

0 and 3

50. myininaya

yes so x-3+3=0+3 gives us x+0=3 Or just x=3

51. myininaya

Now you solve x-9=0

52. myininaya

What are you going to add to both sides?

53. algebra2sucks

9?

54. myininaya

yes so x=?

55. algebra2sucks

9 and 0?

56. algebra2sucks

so x=9

57. myininaya

yes

58. myininaya

Discontinuous at x=3 or x=9

59. myininaya

Now go back to the problem where you had everything factored.

60. myininaya

$\frac{x-3}{(x-3)(x-9)}$

61. myininaya

This was the problem where everything is factored, correct?

62. myininaya

Does anything cancel? If so that is where you have your removable discontinuity.

63. myininaya

x-3 is just a number so is x-9 If we had 5/(5*9) you would say the 5's cancel right? Like this 1/9 Does anything cancel? Is there a factor on top that you see on bottom?

64. myininaya

$\frac{(x-3)}{(x-3)(x-9)}$ If you want you can put parenthesis around the x-3 on top, Do you see any factors that match on top and bottom?

65. myininaya

This is just like matching socks.

66. myininaya

I have to go to the grocery store. Good luck.

67. algebra2sucks

Thanks for all ur help