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Inopeki

  • 4 years ago

TuringTest, what now?

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  1. saifoo.khan
    • 4 years ago
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    Party time! :D

  2. TuringTest
    • 4 years ago
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    lol I dunno...

  3. Inopeki
    • 4 years ago
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    What can you teach me?

  4. TuringTest
    • 4 years ago
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    a lot, there's a fair amount to algebra, but what's the next logical step?

  5. FoolForMath
    • 4 years ago
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    Teach about polynomial.

  6. TuringTest
    • 4 years ago
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    I have already to an extent, but where to go next...

  7. Inopeki
    • 4 years ago
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    Ratios?

  8. FoolForMath
    • 4 years ago
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    In maths the fundamentals concepts are limited but the problems are not, so how about practicing what you have already learned?

  9. TuringTest
    • 4 years ago
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    That's a good point FFM so lets do simplification that you know, but a little more intense: I think I need to make sure you can simplify bigger things:\[\frac{x^2y^5z}{xy^9z^3}\]

  10. saifoo.khan
    • 4 years ago
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  11. TuringTest
    • 4 years ago
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    awww, you're making me think of my cat in the hospital :(

  12. Inopeki
    • 4 years ago
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    How did it go with her?

  13. TuringTest
    • 4 years ago
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    broken leg... waiting for an operation

  14. FoolForMath
    • 4 years ago
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    You beat you cat Turing?!!!

  15. Inopeki
    • 4 years ago
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    aww :(

  16. saifoo.khan
    • 4 years ago
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    turing sat on her leg. :(

  17. TuringTest
    • 4 years ago
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    she fell off the roof. how pessimistic

  18. saifoo.khan
    • 4 years ago
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    awww, sorry i was jking..

  19. TuringTest
    • 4 years ago
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    It's all good :P

  20. FoolForMath
    • 4 years ago
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    Due to Turing's intense physics stress his cat decided to commit suicide :P

  21. Inopeki
    • 4 years ago
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    (x^2)*(y^5)*(z) ------------ = (x)*(y^5/9)*(z^1/3)? (x)*(y^9)*(z^3)

  22. saifoo.khan
    • 4 years ago
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  23. GT
    • 4 years ago
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    Cats are cool.

  24. FoolForMath
    • 4 years ago
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    why \( y^{5/9} \) ?

  25. saifoo.khan
    • 4 years ago
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    no doubt. ;)

  26. TuringTest
    • 4 years ago
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    you have inconsistent rules above inopeki\[\frac{x^a}{x^b}=x^{a-b}\]always...

  27. Inopeki
    • 4 years ago
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    http://i.imgur.com/48wOw.jpg

  28. Inopeki
    • 4 years ago
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    Oh right

  29. Inopeki
    • 4 years ago
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    (x^2)*(y^5)*(z) ------------ = (x)*(y^5-9)*(z^1-3)? (x)*(y^9)*(z^3)

  30. TuringTest
    • 4 years ago
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    yes, simplify...

  31. saifoo.khan
    • 4 years ago
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  32. Inopeki
    • 4 years ago
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    (x)*(y^5-9)*(z^1-3)=(x)*(y^-4)*(z^-2)?

  33. Inopeki
    • 4 years ago
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    http://i.imgur.com/6uzXf.jpg

  34. saifoo.khan
    • 4 years ago
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    cute^

  35. TuringTest
    • 4 years ago
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    yes, do you know another way to write\[x^{-a}\]???

  36. Inopeki
    • 4 years ago
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    No

  37. Inopeki
    • 4 years ago
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    http://i.imgur.com/eqTIb.jpg

  38. TuringTest
    • 4 years ago
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    \[x^{-a}=\frac{1}{x^a}\]so it's probably nicer to rewrite your expression with all positive exponents in this way.

  39. Inopeki
    • 4 years ago
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    Oh right, all negative exponents make the "total number" divided by one.

  40. FoolForMath
    • 4 years ago
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    Btw who can explain why x^0 = 1?

  41. Inopeki
    • 4 years ago
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    (x)*(1/y^4)*(1/z^2)

  42. TuringTest
    • 4 years ago
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    Oh dear... I have my answers about x^0=1 (not true for x=0), but I'm sure FFM would not approve of them :/ @inopeki, yes now rewrite it as 1 fraction...

  43. FoolForMath
    • 4 years ago
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    Turing that's a very important yet fundamental question.

  44. FoolForMath
    • 4 years ago
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    another one is why \( (a^b)^c = a^{bc} \) ?

  45. TuringTest
    • 4 years ago
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    well I have a very simple proof of it and I can show why it is not true for x=0 what more do I need in your opinion?

  46. TuringTest
    • 4 years ago
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    the last rule is easier to explain, perhaps I should...

  47. Inopeki
    • 4 years ago
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    (x)*(1/y^4)*(1/z^2) (x)*(1/y^4*z^2)?

  48. TuringTest
    • 4 years ago
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    yes, now write it as a fraction what goes on the bottom? what goes on the top?

  49. TuringTest
    • 4 years ago
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    @FFMactually that's not so easy to explain now that I think about it.

  50. FoolForMath
    • 4 years ago
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    Can you explain the second rule intuitively?

  51. Inopeki
    • 4 years ago
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    1 x* -------- y^4*z^2

  52. TuringTest
    • 4 years ago
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    no, I was thinking more of how easy it is to show x^ax^b=x^(a+b) intuitively, the other is tricky to me.

  53. TuringTest
    • 4 years ago
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    @ inopeki you can put the x on top, it means the same thing and looks nicer

  54. Inopeki
    • 4 years ago
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    x*1 -------- y^4*z^2

  55. FoolForMath
    • 4 years ago
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    Can we say that all power function obeys the functional equation \( f(x)f(y)= f(x+y) \)?

  56. Inopeki
    • 4 years ago
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    Whats that f?

  57. TuringTest
    • 4 years ago
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    @Inopeki yes, no need to write the 1 though @FFM, I'd have to think about that :/

  58. TuringTest
    • 4 years ago
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    unspecified functions he's asking how far the rule about exponents can be extended..

  59. Inopeki
    • 4 years ago
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    x -------- Oh right y^4*z^2

  60. FoolForMath
    • 4 years ago
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    Actually we can't but I believe that is true for exponential functions though.

  61. Inopeki
    • 4 years ago
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    Umm, ok?

  62. TuringTest
    • 4 years ago
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    it must be for simple ones\[2^x2^y=2^{x+y}\]of course

  63. FoolForMath
    • 4 years ago
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    Tha's whats Inopeki is using right ?

  64. TuringTest
    • 4 years ago
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    basically Inopeki do you see the connection between what you are doing and the rule for multiplying exponents like\[x^ax^b\]?

  65. Inopeki
    • 4 years ago
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    Yeah, but does that mean that it becomes x -------- ? yz^6

  66. TuringTest
    • 4 years ago
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    no because y and z are different bases notice the rule above had both base x.

  67. Inopeki
    • 4 years ago
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    YEah but then i dont see the connection..

  68. TuringTest
    • 4 years ago
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    just that dividing and multiplying are inverse operations x^a=x*x*x*...*x (a times) x^b=x*x*x*...*x (b times) so if we have a=3 b=2 we get x^a --- x^b x*x*x =-----= x x*x which is x^(a-b) if we have (x^a)(x^b) we get (x*x*x)(x*x)=x*x*x*x*x=x^(a+b) so the rules for exponents here come directly from their definitions. You can count the x and see that this relationship holds.

  69. Inopeki
    • 4 years ago
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    I know about that, like x^2*x^8=x^10

  70. Inopeki
    • 4 years ago
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    x^10/x^5=x^5

  71. Inopeki
    • 4 years ago
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    Basic

  72. TuringTest
    • 4 years ago
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    right, I want you to see how that and the rule for division are inverses of each other for a reason...

  73. FoolForMath
    • 4 years ago
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    Turing you have a heck of patience :D

  74. TuringTest
    • 4 years ago
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    eh it's easy FFM, doesn't require too much concentration. good practice too I learn by teaching ;) ok factor \[3x^3y^3+6xy+9x^2y\]

  75. FoolForMath
    • 4 years ago
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    :-)

  76. Inopeki
    • 4 years ago
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    The GCF of 3,6,9 is 3 The GCF of x,x^2,x^3 is x The GCF of y,y^2,y^3 is y 3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Middle step, right?

  77. TuringTest
    • 4 years ago
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    yes exactly

  78. Inopeki
    • 4 years ago
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    3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?

  79. TuringTest
    • 4 years ago
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    yep :) great!

  80. Inopeki
    • 4 years ago
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    Really? :D

  81. FoolForMath
    • 4 years ago
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    Note to Inopeki: Turing is a great resource, learn new things from him and practice as much as you can on your own :-)

  82. TuringTest
    • 4 years ago
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    Yes I've stressed the importance of studying on your own as well and yes Inopeki really, FFM would have noticed a mistake I'm sure ok quick, foil (a-b)(a+b)

  83. Inopeki
    • 4 years ago
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    When hes not here i try to go on purplemath and khan :)

  84. Inopeki
    • 4 years ago
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    a^2+ab-ba-b^2

  85. TuringTest
    • 4 years ago
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    simplify

  86. FoolForMath
    • 4 years ago
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    Books books books!! online learning has it's own limitation :-)

  87. Inopeki
    • 4 years ago
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    a^2-b^2?

  88. Akshay_Budhkar
    • 4 years ago
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    it doesnt @ffm i disagree

  89. Inopeki
    • 4 years ago
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    Foolformath, im having trouble finding books here in sweden

  90. TuringTest
    • 4 years ago
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    and what is the name of that form? remember?

  91. Inopeki
    • 4 years ago
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    The fundamental theorem of algebra? Or want that the one with the multiplex?

  92. TuringTest
    • 4 years ago
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    Multiplex? no I think you're thinking of 'multiplicity' but this is the 'difference of squares' \[a^2-b^2=(a-b)(a+b)\]

  93. Akshay_Budhkar
    • 4 years ago
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    with ocw.mit , khan, purple math and openstudy and people like turing there is no limit to online education ^

  94. TuringTest
    • 4 years ago
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    Thanks but you get what you put in, I think is true with all this stuff.

  95. Inopeki
    • 4 years ago
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    Oh right

  96. TuringTest
    • 4 years ago
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    The fundamental theorem of algebra: "the sum of the multiplicity of the roots of a polynomial is equal to its order" difference of squares\[a^2-b^2=(a-b)(a+b)\]both important, but very different

  97. TuringTest
    • 4 years ago
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    so remember that we can run this backwards, and I can say 'factor'\[x^2-4\]using difference of squares wanna try?

  98. TuringTest
    • 4 years ago
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    run the FOIL backwards I meant*

  99. Inopeki
    • 4 years ago
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    x(x-4)?

  100. Inopeki
    • 4 years ago
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    Oh

  101. TuringTest
    • 4 years ago
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    no look at the form\[a^2-b^2=(a-b)(a+b)\]so what is a and b in\[x^2-4\]???

  102. Inopeki
    • 4 years ago
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    (x-2)(x+2)?

  103. TuringTest
    • 4 years ago
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    there ya go :)

  104. TuringTest
    • 4 years ago
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    how about\[x^4-y^4\](this is a favorite question on OS)

  105. Inopeki
    • 4 years ago
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    (x^2-y^2)(x^2+y^2)

  106. Inopeki
    • 4 years ago
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    Why?

  107. TuringTest
    • 4 years ago
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    good, now is that all we can do with it though? (I've seen a lot of tutors stop here too ;-)

  108. TuringTest
    • 4 years ago
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    hint:look at the first set of parentheses

  109. Inopeki
    • 4 years ago
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    Ummm

  110. TuringTest
    • 4 years ago
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    what is the first sett of parentheses?

  111. Inopeki
    • 4 years ago
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    (x^2-y^2)?

  112. TuringTest
    • 4 years ago
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    yes, and can you factor that?

  113. Inopeki
    • 4 years ago
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    GCF of x^2 is x GCF of y^2 is y xy(x^2/xy)-xy(y^2/xy)?

  114. TuringTest
    • 4 years ago
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    no you're over-thinking x^2-y^2 is difference of squares again!

  115. TuringTest
    • 4 years ago
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    \[a^2-b^2=(a-b)(a+b)\]

  116. Inopeki
    • 4 years ago
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    So (x-y)(x+y)?

  117. TuringTest
    • 4 years ago
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    yes! so now factor\[p^4-q^4\]completely!!

  118. TuringTest
    • 4 years ago
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    you will apply difference of squares twice

  119. Inopeki
    • 4 years ago
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    (p^2-q^2)(p^2+q^2) (p-q)(p+q)(p+q)+(p+q)?

  120. Inopeki
    • 4 years ago
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    Without the +

  121. Inopeki
    • 4 years ago
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    between the parenthesis

  122. TuringTest
    • 4 years ago
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    (p^2-q^2)(p^2+q^2) stop here we cannot factor the second set of parentheses, it is not a /difference/ of squares, it is a /sum/ of squares, and we have no formula for that. check that (a+b)(a+b) is not a^2+b^2 so only factor the one that is a /difference/ of squares

  123. Inopeki
    • 4 years ago
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    (p-q)(p+q)(p^2+q^2) You said i had to do difference in squares 2 times? Btw, lets move to a new thread, this one is starting to lag a little bit

  124. TuringTest
    • 4 years ago
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    That is above correct, you did difference of squares twice: p^4-q^4 (p^2-q^2)(p^2+q^2) <-once (p-q)(p+q)(p^2+q^2) <-twice ok new thread...

  125. Inopeki
    • 4 years ago
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    Ohhh

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