anonymous one year ago (y+3)^3+8 ? how to factor completely ?

1. UsukiDoll

sum of cubes formula $(a+b)^3 =(a+b)(a^2-ab+b^2)$ rewrite $(y+3)^3=8$ as $(y+3)^3=2^3$ so let a = y+3 and b = 2

2. anonymous

hmm... then ?:(

3. UsukiDoll

well the b portion is easy because if b = 2 then b^2 is ?

4. UsukiDoll

OH MAN! I GOT MY LATEX WRONG NUGH! $\large (y+3)^3+ 8 \rightarrow = (y+3)^3+2^3$

5. UsukiDoll

but the sum of cubes formula is correct thank goodness.

6. UsukiDoll

so if b = 2 what's b^2 = ?

7. anonymous

4?

8. UsukiDoll

$\large (y+3)^3+ 8 \rightarrow (y+3)^3+2^3$ idk what's with me in the middle of the night

9. UsukiDoll

but yes b^2 = 4 because b = 2 2^2 = 4 woo.

10. UsukiDoll

$((y+3)+(2))^3 =((y+3)+2)((y+3)^2-2(y+3)+4)$

11. anonymous

(y+3+2) = (y+5) (_______)

12. UsukiDoll

so we need to expand (y+3)^2 and distribute the -2 on y+3

13. anonymous

(y+5)(y^2+9-(y+3+2)+4) ?

14. UsukiDoll

wait.. try expand this in parts expand $(y+3)^2$ and then distribute -2(y+3)

15. imqwerty

i didn't even say something :(

16. anonymous

y^2+3y+9 then?

17. imqwerty

i ws here to help k..sry bye i won't interrupt u frm nw .

18. UsukiDoll

then distribute the -2 for -2(y+3)

19. anonymous

-2y-6

20. UsukiDoll

yes so now we can place them into our equation $((y+3)+(2))^3 =((y+3)+2)((y+3)^2-2(y+3)+4)$ $((y+3)+(2))^3 =(y+5)(y^2+3y+9-2y-6+4)$

21. UsukiDoll

then we can rearrange and combine like terms...

22. UsukiDoll

$(y+5)(y^2+3y-2y-6+4+9)$

23. UsukiDoll

so combine like terms in the super long equation what's 3y-2y what's -6+4+9

24. anonymous

(y+5)(y^2+y+7) ?

25. UsukiDoll

cool so now we need to ... wow distribute $(y+5)(y^2+y+7)$ so distribute the y throughout $y^2+y+7$ like $y(y^2+y+7)$

26. UsukiDoll

$(y^2(y)+y(y)+7(y))$

27. UsukiDoll

do you know the exponent laws ?

28. anonymous

yup

29. UsukiDoll

alrighty then :)

30. anonymous

y^3+y^2+7y

31. UsukiDoll

excellent! now we do the same thing for the 5 $(y+5)(y^2+y+7)$ so distribute 5 throughout $y^2+y+7$ $5(y^2+y+7)$

32. UsukiDoll

so it's like 5 x y^2 5 x y 5 x 7

33. anonymous

5y^2 +5y+35

34. UsukiDoll

yeah! so now we combine like terms $y^3+y^2+7y+5y^2+5y+35$

35. UsukiDoll

-_- something is wrong ... our middle terms are off.

36. anonymous

y^3+5y^2+12y+35

37. UsukiDoll

yeah but I'm double checking through a calculator and it claims that it's 9y^2+27y unless ... hey @Haseeb96 could you come over here and check this?

38. UsukiDoll

oh nevermind.. wait let me think... our first term and last term is correct..

39. anonymous

i hate factoring completely :(

40. anonymous

bye :) i have lot to do :/

41. UsukiDoll

ok. I'll try and figure out what's up before I go sleep. Maybe we needed pascal's triangle/binomial theorem.

42. anonymous

thanks for your time dude :)

43. UsukiDoll

you're welcome. I'm gonna post the pascal's triangle version.. I see it clearer on how they got the middle term... another method is to exapand (y+3)^3 manually and add 8 but who wants to do that ? that's time consuming.

44. UsukiDoll

|dw:1437311220374:dw|

45. UsukiDoll

so from k = 0 to k = 3 ...

46. UsukiDoll

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47. UsukiDoll

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48. UsukiDoll

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49. UsukiDoll

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50. UsukiDoll

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51. UsukiDoll

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52. UsukiDoll

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53. UsukiDoll

@freckles wait don't go! I was using sum of cubes and the middle terms were wonky . What gives?

54. UsukiDoll

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55. UsukiDoll

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56. UsukiDoll

ugh this thing is a troll x.x

57. UsukiDoll

y^3+9y^2+27y+35 ... should be it but the middle terms splash out when I did sum of cubes. So I'm doing Pascal's triangle / Binomial Theorem and that's working so far

58. UsukiDoll

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59. freckles

$(y+3)^3+8 \\ a^3+b^3=(a+b)(a^2-ab+b^2) \\ a=y+3 \\ b=2 \\ a^2=(y+3)^2 \\ b^2=4 \\ a^3=(y+3)^3 \\ b^3=8 \\ \\ \text{ input into formula } \\ (y+3)^3+8=(y+3+2)([y+3]^2-[y+3](2)+4) \\ (y+3)^3+8=(y+5)(y^2+6y+9-2y-6+4) \\ \text{ combine like terms inside the second () }$

60. UsukiDoll

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61. UsukiDoll

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62. UsukiDoll

|dw:1437311906118:dw| that should be 1 since everything cancels.

63. UsukiDoll

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64. freckles

I think I see what you did wrong

65. freckles

$(y+3)^2=y^2+6y+9 \\ (y+3)^2 \neq y^2+3y+9$

66. UsukiDoll

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67. UsukiDoll

OMg x.x Not my night

68. freckles

it was probably just a type-o and you just never fixed and you forgot about it or whatever happens to me too

69. UsukiDoll

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70. UsukiDoll

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71. UsukiDoll

oh yay

72. UsukiDoll

$(y+5)(y^2+4y+7)$ $y^3+4y^2+7y+5y^2+20y+35 = y^3+9y^2+27y+35$

73. UsukiDoll

there we go

74. UsukiDoll

because (y+3)(y+3) = y^2+3x+3x+9 = y^2+6x+9 I must have overlooked it. so sum of cubes can be used after all.

75. UsukiDoll

@freckles I got it to work :)