## anonymous one year ago okay help with this problem and then one more and I wont need anymore help tonight. (2-3i(3-i)-(4-i)(4+i)

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1. anonymous

$(2-3i)(3-i)-(4-i)(4+i)$

2. anonymous

(2-3i)(3-i)-(4-i)(4+i)

3. anonymous

theres supposed to be one more parenthases after 3i

4. anonymous

5. jim_thompson5910

you can FOIL or use the table like last time. Which option works best for you?

6. anonymous

foil the table is a little confusing.

7. jim_thompson5910

ok what do you get when you FOIL (2-3i)(3-i)

(2 -3i)(3-i)-(4-i)(4+i) (2)(3-i)-(3i)(3-i)-(4^2-i^2) Multiply find the value Note i^2 = -1 Source: mathskey.com

9. anonymous

3-11i

10. jim_thompson5910

correct

11. jim_thompson5910

how about (4-i)(4+i) ?

12. anonymous

that is 15

13. anonymous

brb

14. jim_thompson5910

not 15, but you are close

15. jim_thompson5910

try again

16. anonymous

okay

17. anonymous

where did i go wrong with my problem i will show you my work.

(4-i)(4+i) (a-b)(a+b) =a^2-b^2 4^2 -i^2 16-(-1)=17

19. anonymous

20. anonymous

|dw:1441072256502:dw|

21. jim_thompson5910

i^2 is -1 so $$\Large 16-\color{red}{i^2}$$ turns into $$\Large 16-\color{red}{(-1)}$$

22. anonymous

im not sure @sayomi

23. jim_thompson5910

Do you see what I mean @huntergirl1 ?

24. anonymous

yes now i do @jim_thompson5910

25. jim_thompson5910

ok great, so 16-(-1) = 16+1 = 17 --------------------- $\large \color{red}{(2-3i)(3-i)}-\color{blue}{(4-i)(4+i)} =\color{red}{3-11i}-\color{blue}{17} = -14-11i$

26. anonymous

@sayomi you were correct and yes @jim_thompson5910 i do understand

27. jim_thompson5910

28. anonymous

so the ending result is 51-187i

29. anonymous

okay do you wnt me to start a new window for the next problem? or to place it here?

30. jim_thompson5910

how are you getting 51-187i ?

31. jim_thompson5910

32. anonymous

bc you put 17 back into the problem

33. jim_thompson5910

are you saying that they wanted you to multiply (2-3i)(3-i) and (4-i)(4+i) ??

34. anonymous

actually i put those answers both into the website they were both incorrect

35. anonymous

hwy was the answer supposed to be -14-11i?

36. anonymous

why

37. jim_thompson5910

because the 3 and -17 combine to -14

38. jim_thompson5910

(3-11i) - (17) = 3-11i - 17 = (3-17) + (-11i) = -14 - 11i

39. anonymous

oh okay.

40. anonymous

will you be on to check a few more problems before I submit them?

41. jim_thompson5910

sure

42. anonymous

thank you!!!!!!!!!!!!

43. jim_thompson5910

no problem

44. anonymous

okay i have one to check its similar to the last problem

45. anonymous

|dw:1441074082197:dw|

46. anonymous

what did i do wrong?

47. jim_thompson5910

you didn't foil correctly

48. anonymous

okay i guess the sqrt part mixed me up

49. anonymous

i also tried the box method too

50. jim_thompson5910

|dw:1441074547608:dw|

51. jim_thompson5910

what we do is make a 2x2 box 2 rows and 2 columns the terms of each factor will go along the outside like this |dw:1441074595637:dw|

52. jim_thompson5910

agreed so far?

53. anonymous

okay thats how i started it on my paper too

54. jim_thompson5910

ok go ahead and fill it out

55. anonymous

|dw:1441074610390:dw|

56. jim_thompson5910

-4 times $$\large -i\sqrt{2}$$ should be $$\large +4i\sqrt{2}$$ the two negatives cancel |dw:1441074917203:dw|

57. anonymous

right?

58. jim_thompson5910

same with the -i and -i being multiplied in the lower right box |dw:1441074956036:dw|

59. anonymous

so instead of a negative is a possitive

60. jim_thompson5910

so after filling the boxes and simplifying the expressions in each box, we end up with |dw:1441075014939:dw| since $\Large i^2*\sqrt{4} = -1*2 = -2$

61. anonymous

okay

62. anonymous

63. anonymous

64. jim_thompson5910

now add up the terms inside the boxes $\Large 16 + 4i\sqrt{2} + 4i\sqrt{2}+(-2) = 14+8i\sqrt{2}$

65. jim_thompson5910

|dw:1441075415932:dw|

66. anonymous

okay that makes since

67. jim_thompson5910

|dw:1441075435745:dw|