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anonymous

  • one year ago

(y+3)^3+8 ? how to factor completely ?

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  1. UsukiDoll
    • one year ago
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    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
    • one year ago
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    hmm... then ?:(

  3. UsukiDoll
    • one year ago
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    well the b portion is easy because if b = 2 then b^2 is ?

  4. UsukiDoll
    • one year ago
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    OH MAN! I GOT MY LATEX WRONG NUGH! \[\large (y+3)^3+ 8 \rightarrow = (y+3)^3+2^3 \]

  5. UsukiDoll
    • one year ago
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    but the sum of cubes formula is correct thank goodness.

  6. UsukiDoll
    • one year ago
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    so if b = 2 what's b^2 = ?

  7. anonymous
    • one year ago
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    4?

  8. UsukiDoll
    • one year ago
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    \[\large (y+3)^3+ 8 \rightarrow (y+3)^3+2^3\] idk what's with me in the middle of the night

  9. UsukiDoll
    • one year ago
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    but yes b^2 = 4 because b = 2 2^2 = 4 woo.

  10. UsukiDoll
    • one year ago
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    \[((y+3)+(2))^3 =((y+3)+2)((y+3)^2-2(y+3)+4)\]

  11. anonymous
    • one year ago
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    (y+3+2) = (y+5) (_______)

  12. UsukiDoll
    • one year ago
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    so we need to expand (y+3)^2 and distribute the -2 on y+3

  13. anonymous
    • one year ago
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    (y+5)(y^2+9-(y+3+2)+4) ?

  14. UsukiDoll
    • one year ago
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    wait.. try expand this in parts expand \[(y+3)^2 \] and then distribute -2(y+3)

  15. imqwerty
    • one year ago
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    i didn't even say something :(

  16. anonymous
    • one year ago
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    y^2+3y+9 then?

  17. imqwerty
    • one year ago
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    i ws here to help k..sry bye i won't interrupt u frm nw .

  18. UsukiDoll
    • one year ago
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    then distribute the -2 for -2(y+3)

  19. anonymous
    • one year ago
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    -2y-6

  20. UsukiDoll
    • one year ago
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    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
    • one year ago
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    then we can rearrange and combine like terms...

  22. UsukiDoll
    • one year ago
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    \[(y+5)(y^2+3y-2y-6+4+9)\]

  23. UsukiDoll
    • one year ago
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    so combine like terms in the super long equation what's 3y-2y what's -6+4+9

  24. anonymous
    • one year ago
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    (y+5)(y^2+y+7) ?

  25. UsukiDoll
    • one year ago
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    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
    • one year ago
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    \[(y^2(y)+y(y)+7(y))\]

  27. UsukiDoll
    • one year ago
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    do you know the exponent laws ?

  28. anonymous
    • one year ago
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    yup

  29. UsukiDoll
    • one year ago
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    alrighty then :)

  30. anonymous
    • one year ago
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    y^3+y^2+7y

  31. UsukiDoll
    • one year ago
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    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
    • one year ago
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    so it's like 5 x y^2 5 x y 5 x 7

  33. anonymous
    • one year ago
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    5y^2 +5y+35

  34. UsukiDoll
    • one year ago
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    yeah! so now we combine like terms \[y^3+y^2+7y+5y^2+5y+35\]

  35. UsukiDoll
    • one year ago
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    -_- something is wrong ... our middle terms are off.

  36. anonymous
    • one year ago
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    y^3+5y^2+12y+35

  37. UsukiDoll
    • one year ago
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    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
    • one year ago
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    oh nevermind.. wait let me think... our first term and last term is correct..

  39. anonymous
    • one year ago
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    i hate factoring completely :(

  40. anonymous
    • one year ago
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    bye :) i have lot to do :/

  41. UsukiDoll
    • one year ago
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    ok. I'll try and figure out what's up before I go sleep. Maybe we needed pascal's triangle/binomial theorem.

  42. anonymous
    • one year ago
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    thanks for your time dude :)

  43. UsukiDoll
    • one year ago
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    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
    • one year ago
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    |dw:1437311220374:dw|

  45. UsukiDoll
    • one year ago
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    so from k = 0 to k = 3 ...

  46. UsukiDoll
    • one year ago
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    |dw:1437311295892:dw|

  47. UsukiDoll
    • one year ago
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    |dw:1437311393568:dw|

  48. UsukiDoll
    • one year ago
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    |dw:1437311428990:dw|

  49. UsukiDoll
    • one year ago
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    |dw:1437311489980:dw|

  50. UsukiDoll
    • one year ago
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    |dw:1437311527971:dw|

  51. UsukiDoll
    • one year ago
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    |dw:1437311547143:dw|

  52. UsukiDoll
    • one year ago
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    |dw:1437311614971:dw|

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

  54. UsukiDoll
    • one year ago
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    |dw:1437311655716:dw|

  55. UsukiDoll
    • one year ago
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    |dw:1437311666239:dw|

  56. UsukiDoll
    • one year ago
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    ugh this thing is a troll x.x

  57. UsukiDoll
    • one year ago
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    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
    • one year ago
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    |dw:1437311833243:dw|

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

  60. UsukiDoll
    • one year ago
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    |dw:1437311848473:dw|

  61. UsukiDoll
    • one year ago
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    |dw:1437311880098:dw|

  62. UsukiDoll
    • one year ago
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    |dw:1437311906118:dw| that should be 1 since everything cancels.

  63. UsukiDoll
    • one year ago
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    |dw:1437311933179:dw|

  64. freckles
    • one year ago
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    I think I see what you did wrong

  65. freckles
    • one year ago
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    \[(y+3)^2=y^2+6y+9 \\ (y+3)^2 \neq y^2+3y+9\]

  66. UsukiDoll
    • one year ago
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    |dw:1437311941673:dw|

  67. UsukiDoll
    • one year ago
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    OMg x.x Not my night

  68. freckles
    • one year ago
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    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
    • one year ago
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    |dw:1437312099694:dw|

  70. UsukiDoll
    • one year ago
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    |dw:1437312142306:dw|

  71. UsukiDoll
    • one year ago
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    oh yay

  72. UsukiDoll
    • one year ago
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    \[ (y+5)(y^2+4y+7)\] \[y^3+4y^2+7y+5y^2+20y+35 = y^3+9y^2+27y+35\]

  73. UsukiDoll
    • one year ago
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    there we go

  74. UsukiDoll
    • one year ago
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    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
    • one year ago
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    @freckles I got it to work :)

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