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anonymous

  • 5 years ago

i have a weird question maybe someone can help.... q^2+1...how do u factor?

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  1. anonymous
    • 5 years ago
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    you cannot with real numbers

  2. anonymous
    • 5 years ago
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    That's true..

  3. anonymous
    • 5 years ago
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    so does that mean its prime?

  4. anonymous
    • 5 years ago
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    \[(q-1)(q-1)\]

  5. anonymous
    • 5 years ago
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    that is not correct

  6. anonymous
    • 5 years ago
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    No, if you are to solve for q, you have \[\sqrt{-1}=\sqrt{i}\] This requires that you are familiar with complex numbers.

  7. anonymous
    • 5 years ago
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    but then the middle would be -2q

  8. anonymous
    • 5 years ago
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    \[(q+1)(q+1)\]

  9. anonymous
    • 5 years ago
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    then be +2

  10. anonymous
    • 5 years ago
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    i mean +2q

  11. myininaya
    • 5 years ago
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    does (q-i)(q+i) work

  12. anonymous
    • 5 years ago
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    no

  13. anonymous
    • 5 years ago
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    no cause then it would be -1 myininaya

  14. myininaya
    • 5 years ago
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    q^2+iq-iq-i^2 q^2+0-(-1) q^2+1 yes it works ! :)

  15. anonymous
    • 5 years ago
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    but its q^2+1

  16. anonymous
    • 5 years ago
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    listen

  17. anonymous
    • 5 years ago
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    i=sqrt(-1)

  18. myininaya
    • 5 years ago
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    q^2+1=(q-i)(q+i)

  19. anonymous
    • 5 years ago
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    that is called a complex number

  20. anonymous
    • 5 years ago
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    has a lot of interesting bits

  21. anonymous
    • 5 years ago
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    -1*+1 is still -1

  22. anonymous
    • 5 years ago
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    nooooooooooooooooooooooooooooooooooooooooo

  23. myininaya
    • 5 years ago
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    (q-i)(q+i)=q^2-qi+qi-i^2 agreed?

  24. anonymous
    • 5 years ago
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    -i*i=-i^2=-(-1)=1

  25. anonymous
    • 5 years ago
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    its not i its 1(one)

  26. myininaya
    • 5 years ago
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    q^2+0-(-1) q^2+1

  27. myininaya
    • 5 years ago
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    i^2=-1

  28. anonymous
    • 5 years ago
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    \[(q+1)(q+1) = q ^{2}+1^{2}\]

  29. myininaya
    • 5 years ago
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    lol this is correct to say q^2+1=(q-i)(q+i)

  30. anonymous
    • 5 years ago
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    NOOOOO it is not

  31. anonymous
    • 5 years ago
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    either im super stupid or im just not comprehending how that works

  32. myininaya
    • 5 years ago
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    shark thats wrong (q+1)(q+1)=q^2+2q+1

  33. anonymous
    • 5 years ago
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    sharkattack you still get q^2 2q+1 as the answer

  34. anonymous
    • 5 years ago
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    As I said: you cannot with real numbers! (you need to introduce complex numbers with i to solve it)

  35. anonymous
    • 5 years ago
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    so andras does that mean its prime?

  36. anonymous
    • 5 years ago
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    \[(X+Y)(X+Y) = X ^{2} + Y ^{2}\]

  37. anonymous
    • 5 years ago
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    no it does not

  38. myininaya
    • 5 years ago
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    ok Bmartin i^2=-1 (q-i)(q+i)=q^2+qi-qi-i^2 =q^2-(-1) =q^2+1

  39. anonymous
    • 5 years ago
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    for example 3^2+1=10

  40. myininaya
    • 5 years ago
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    bmartin do you see the negative in front of the i^2?

  41. anonymous
    • 5 years ago
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    not prime

  42. anonymous
    • 5 years ago
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    so -1*+1 =+1?

  43. anonymous
    • 5 years ago
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    i dont believe that

  44. anonymous
    • 5 years ago
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    sorry but I'm right

  45. myininaya
    • 5 years ago
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    -(-1)=1

  46. myininaya
    • 5 years ago
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    i^2=-1 not 1

  47. anonymous
    • 5 years ago
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    myiniaya you are just confusing them with complex numbers

  48. myininaya
    • 5 years ago
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    you can factor over complex numbers

  49. anonymous
    • 5 years ago
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    Sharkattack is right!!! Lets vote him for president!!!

  50. anonymous
    • 5 years ago
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    lol

  51. anonymous
    • 5 years ago
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    :-)

  52. anonymous
    • 5 years ago
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    so the answer is (q+1)(q-1)? i just really dont see how that works....

  53. myininaya
    • 5 years ago
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    no! it is (q-i)(q+i)

  54. anonymous
    • 5 years ago
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    :)

  55. anonymous
    • 5 years ago
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    no there is no answer

  56. anonymous
    • 5 years ago
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    why does it make a difference where the signs go?

  57. anonymous
    • 5 years ago
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    no

  58. anonymous
    • 5 years ago
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    with normal (real) numbers there is no answer

  59. anonymous
    • 5 years ago
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    oooooooooooooh im soooooo confused

  60. anonymous
    • 5 years ago
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    I can imagine, what is not clear?

  61. anonymous
    • 5 years ago
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    \[(X+Y)(X+Y) =X ^{2} +Y ^{2}\]

  62. myininaya
    • 5 years ago
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    lol (q-i)(q+i)=q^2-qi+qi-i^2=q^2-(-1)=q^2+1 how is not one seeing this except me i dont want to be on a debate team accept my answer or not you can either say not factorable over the real numbers or you can say the answer is (q-i)(q+i)

  63. anonymous
    • 5 years ago
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    shark it is not: it is x^2+2xy+y^2

  64. anonymous
    • 5 years ago
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    \[(q+1)(q+1) =q ^{2} +1^{2} =q ^{2} +1\]

  65. anonymous
    • 5 years ago
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    i just keep seeing that answer as q^2-1 myininaya

  66. anonymous
    • 5 years ago
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    myininaya you are right but if someone never learnt about complex numbers than you cannot explain them in 2 min

  67. myininaya
    • 5 years ago
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    i^2=-1 -i^2=1

  68. anonymous
    • 5 years ago
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    they have the same base

  69. myininaya
    • 5 years ago
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    do you know the bmartin that -i^2=1?

  70. anonymous
    • 5 years ago
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    sharck when you expand a bracket you have to multiply every term with every term so (x+1)(x+1)=x^2+x+x+1

  71. anonymous
    • 5 years ago
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    mmmmm....

  72. myininaya
    • 5 years ago
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    bmartin, do you know that -i^2=1?

  73. anonymous
    • 5 years ago
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    no

  74. myininaya
    • 5 years ago
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    do you about imaginary numberts>

  75. anonymous
    • 5 years ago
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    no

  76. anonymous
    • 5 years ago
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    i'm right in my law

  77. myininaya
    • 5 years ago
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    ok the sum of two real squares is not factorable over the reals

  78. anonymous
    • 5 years ago
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    wait

  79. anonymous
    • 5 years ago
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    ok

  80. myininaya
    • 5 years ago
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    if you knew of imaginary numbers then we could factor it

  81. anonymous
    • 5 years ago
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    factorize x^2 +1 X^2+2xy+Y^2

  82. anonymous
    • 5 years ago
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    hopefully i get to that chapter soon

  83. myininaya
    • 5 years ago
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    lol sorry andras i thought they were talking about imagary numbers so i just figured they knew lol

  84. anonymous
    • 5 years ago
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    no prob

  85. anonymous
    • 5 years ago
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    well thanks everyone!

  86. anonymous
    • 5 years ago
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    this took longer than I thaught

  87. anonymous
    • 5 years ago
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    but

  88. myininaya
    • 5 years ago
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    by the way sean the square root of -1 is i not the square root of i

  89. anonymous
    • 5 years ago
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    yeah, true :-)

  90. anonymous
    • 5 years ago
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    I'm right

  91. anonymous
    • 5 years ago
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    lol

  92. myininaya
    • 5 years ago
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    shark is so cute

  93. anonymous
    • 5 years ago
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    you are as always, go for politician you would do great

  94. anonymous
    • 5 years ago
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    so persistant...you shoud be a lawyer

  95. myininaya
    • 5 years ago
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    i felt as i was debating against someone such tough debater

  96. anonymous
    • 5 years ago
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    i bet he was on the debate team!

  97. anonymous
    • 5 years ago
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    can we convince you shark that you are wrong?

  98. myininaya
    • 5 years ago
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    what about if we 9^2+1 is this equal to 9^2+9(1)+1^2

  99. myininaya
    • 5 years ago
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    oops i mean 9^2+2(9)(1)+1^2

  100. myininaya
    • 5 years ago
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    say no lol

  101. anonymous
    • 5 years ago
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    no

  102. anonymous
    • 5 years ago
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    look

  103. anonymous
    • 5 years ago
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    ha ha well yall have a good night!! thanks for your help! ill beback for more debate later!

  104. myininaya
    • 5 years ago
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    ok im looking

  105. anonymous
    • 5 years ago
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    factorize (x+y)^2 and X^2+Y^2

  106. myininaya
    • 5 years ago
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    what do you mean? (x+y)^2 is already in factorable form x^2+y^2 cannot be factored over the reals

  107. anonymous
    • 5 years ago
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    no

  108. anonymous
    • 5 years ago
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    what's your work ?

  109. myininaya
    • 5 years ago
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    let x=3 and y=2 so (x+y)^2=(3+2)^2=5^2=25 and x^2+y^2=3^2+2^2=9+4=13 13 isn't the same as 26 so (x+y)^2 is not the same as x^2+y^2

  110. myininaya
    • 5 years ago
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    13 isn't the same as 25*

  111. anonymous
    • 5 years ago
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    they are have different factorization

  112. myininaya
    • 5 years ago
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    they are not them same i just gave you a counterexample

  113. anonymous
    • 5 years ago
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    shark you are the best!

  114. myininaya
    • 5 years ago
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    lol

  115. anonymous
    • 5 years ago
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    answer me

  116. anonymous
    • 5 years ago
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    factorize (X+Y)^2 and X^2 + Y^2

  117. myininaya
    • 5 years ago
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    if i multiply (x+y)^2=(x+y)(x+y)=x^2+2xy+y^2 x^2+y^2 is not equal to that since it is missing the 2xy term

  118. anonymous
    • 5 years ago
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    what's your work

  119. myininaya
    • 5 years ago
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    (x+y)(x+y)=x^2+xy+xy+y^2=x^2+2xy+y^2 see the 2xy 2xy is not in x^2+y^2 so (x+y)^2 does not equal x^2+y^2

  120. anonymous
    • 5 years ago
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    i know bec. (X+Y)^2 = X^2 +2xy +Y^2 but X^2 +Y^2 = (X+Y) (X+Y) or (X-Y) (X-Y) factorize X^2 -Y^2

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