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anonymous

  • 5 years ago

Please help! Show me how to factor these binomials! m^4-256 and z^3+125, thank you! Descriptions are appreciated :)

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  1. anonymous
    • 5 years ago
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    m^4-256 this is a perfect square

  2. anonymous
    • 5 years ago
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    because 2*2=4, right?

  3. anonymous
    • 5 years ago
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    the 1 you want to concentrate on is 256

  4. anonymous
    • 5 years ago
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    ok

  5. asnaseer
    • 5 years ago
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    try and get it into the form \(a^2-b^2\) Use the hints that mth3v4 has given you.

  6. anonymous
    • 5 years ago
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    cool pic asna :D

  7. asnaseer
    • 5 years ago
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    thx :) I'm actually an alien from another planet :D

  8. anonymous
    • 5 years ago
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    :)

  9. asnaseer
    • 5 years ago
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    Starshine7 - can you express 256 as (something)^2 ?

  10. anonymous
    • 5 years ago
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    16x16?

  11. anonymous
    • 5 years ago
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    :)

  12. asnaseer
    • 5 years ago
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    good, now can you express \(m^4\) as (something)^2 ?

  13. anonymous
    • 5 years ago
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    so can you put that in in a factored form

  14. anonymous
    • 5 years ago
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    m*m*m*m?

  15. asnaseer
    • 5 years ago
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    Yes, so \(m^4=(m^2)^2\) Now use those two to get the expression into the form \(a^2-b^2\) so that you can then factor it like this:\[a^2-b^2=(a+b)(a-b)\]

  16. anonymous
    • 5 years ago
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    why is it written like (m+number)(trinomial)? final answer i mean. If it helps, I factor by grouping if i can incorporate that somehow

  17. asnaseer
    • 5 years ago
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    just do this one step at a time and al will be revealed :)

  18. anonymous
    • 5 years ago
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    ok

  19. anonymous
    • 5 years ago
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    lol oops

  20. anonymous
    • 5 years ago
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    nothing happened

  21. asnaseer
    • 5 years ago
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    :)

  22. asnaseer
    • 5 years ago
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    Starshine7 - are you stuck?

  23. anonymous
    • 5 years ago
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    Sadly yes :(

  24. anonymous
    • 5 years ago
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    look at the first answer you gave me

  25. anonymous
    • 5 years ago
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    how did you come up with that ?

  26. asnaseer
    • 5 years ago
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    ok, we saw above that:\[256=16^2\]\[m^4=(m^2)^2\]so now we can rewrite your original equation as:\[m^4-256=(m^2)^2-16^2\]now mak use of:\[a^2-b^2=(a+b)(a-b)\]to simplify this further.

  27. anonymous
    • 5 years ago
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    after that if can be factored again as an option as you can see

  28. anonymous
    • 5 years ago
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    (m^2)^2 confuses me

  29. anonymous
    • 5 years ago
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    (i think i am confusing starshine)

  30. asnaseer
    • 5 years ago
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    ok, lets use a substitution to make it clearer...

  31. asnaseer
    • 5 years ago
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    let \(n=m^2\), then we have:\[m^4-256=(m^2)^2-16^2=n^2-16^2\]

  32. anonymous
    • 5 years ago
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    why n? o.O

  33. asnaseer
    • 5 years ago
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    we could use any other symbol - it doesn't really matter

  34. asnaseer
    • 5 years ago
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    the point is it is now in the form \(a^2-b^2\)

  35. asnaseer
    • 5 years ago
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    so you should now be able to factorise it.

  36. anonymous
    • 5 years ago
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    okies properties of exponents state when you multiply exponent and exponent you add the exponent values together (a^x)(a^b)=a^x+b

  37. anonymous
    • 5 years ago
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    exponent properties/laws of exponents whatever

  38. asnaseer
    • 5 years ago
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    Starshine7 - are you familiar with the exponent laws mth3v4 has shown? i.e.:\[(x^a)^b=x^{ab}\]

  39. anonymous
    • 5 years ago
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    If it helps I dont understand when a and b are examples, I was taught using only the numbers, so thats why im really confused :(

  40. anonymous
    • 5 years ago
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    those are just any number

  41. anonymous
    • 5 years ago
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    I know >_> Im weird like that

  42. asnaseer
    • 5 years ago
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    so you are comfortable with something like this:\[18^2-16^2=(18+16)(18-16)\]but when replaced with symbols like this:\[a^2-b^2=(a+b)(a-b)\]you get lost?

  43. anonymous
    • 5 years ago
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    Yes, exactly :)

  44. asnaseer
    • 5 years ago
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    ok - the 'a' and 'b' are just placeholders for any number

  45. anonymous
    • 5 years ago
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    I just have attention issues so the numbers help me understand it better, i drift easily.

  46. asnaseer
    • 5 years ago
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    what this says is that the formula:\[a^2-b^2=(a+b)(a-b)\]is true for ANY value of 'a' and 'b'

  47. asnaseer
    • 5 years ago
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    so in your example above, we ended up with:\[m^4-256=(m^2)^2-16^2=n^2-16^2\]which means we can think of it as a=n and b=16 to get:\[n^2-16^2=(n+16)(n-16)\]we can then substitute back \(n=m^2\) to get:\[m^4-256=(n+16)(n-16)=(m^2+16)(m^2-16)\]

  48. asnaseer
    • 5 years ago
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    do you understand those steps?

  49. anonymous
    • 5 years ago
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    I think so

  50. anonymous
    • 5 years ago
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    (a ^x ) ( a ^b )= a^x+b ^ ^ ^ ^ ^ ^ ^ | | | | / | | | L _____|___ L _ |___L_ _ L _ exponent number both x and b can be any number | | | your number ______ |

  51. anonymous
    • 5 years ago
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    does my funny drawing help XD

  52. asnaseer
    • 5 years ago
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    ok, so now you should notice that we have \((m^2-16)\) as one of the factors. since \(16=4^2\), we can write this as:\[(m^2-16)=(m^2-4^2)=(m+4)(m-4)\]using the same formula. we therefore end up with:\[m^4-256=(m^2+16)(m+4)(m-4)\]

  53. anonymous
    • 5 years ago
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    So you just keep factoring to the lowest numbers that are perfect squares?

  54. asnaseer
    • 5 years ago
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    yes

  55. asnaseer
    • 5 years ago
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    mth3v4 - you have a lot of patience to draw all that :)

  56. anonymous
    • 5 years ago
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    lol

  57. anonymous
    • 5 years ago
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    :)

  58. anonymous
    • 5 years ago
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    it still comes out a bit messy

  59. anonymous
    • 5 years ago
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    (a^x) (a^y) = a ^ x+y lemme try this you have this thing here right starshine?

  60. asnaseer
    • 5 years ago
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    Starshine7 - for your second expression, you need to use the following factorisation:\[a^3+b^3=(a+b)(a^2-ab+b^2)\]so you need to get your expression:\[z^3+125\]into the form \(a^3+b^3\)

  61. anonymous
    • 5 years ago
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    okies finish what you have first

  62. anonymous
    • 5 years ago
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    I guess

  63. asnaseer
    • 5 years ago
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    I have to go now guys - Starshine I am sure mth3v4 will help you with the rest. Good luck all...

  64. anonymous
    • 5 years ago
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    Aww..well bye Asna :) thanks anyways

  65. anonymous
    • 5 years ago
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    kk bb

  66. asnaseer
    • 5 years ago
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    yw

  67. anonymous
    • 5 years ago
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    z^3+125 is your next 1

  68. anonymous
    • 5 years ago
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    Yes

  69. anonymous
    • 5 years ago
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    it is not going to be squared but cubed

  70. anonymous
    • 5 years ago
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    have you heard of diff of cubes or sum of cubes formulas

  71. anonymous
    • 5 years ago
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    no

  72. anonymous
    • 5 years ago
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    so what ansa wrote there before leaving was formula for diff of cubes

  73. anonymous
    • 5 years ago
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    because you just see just a 2 termed expression here

  74. anonymous
    • 5 years ago
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    so how would you make x^3 and change it to x^2

  75. anonymous
    • 5 years ago
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    better yet lemme ask the question another way how do you get x^3 from x^2

  76. anonymous
    • 5 years ago
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    I'm not sure. I think they only have 2 exs in common

  77. anonymous
    • 5 years ago
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    okies lets try this ( a^x ) ( a^y ) = a ^ x +y the variable "a" can be any number 1, 2, 3, whatever, happy face , i dont care what you want it to be ok so far

  78. anonymous
    • 5 years ago
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    this just goes fot the variable "a" we good?

  79. anonymous
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    yes

  80. anonymous
    • 5 years ago
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    then next this for the variables "x, y" these give a numeric value (these have to be a number) giving a value ( a^x ) ( a^y) = a ^ x + y

  81. anonymous
    • 5 years ago
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    so in the end the numeric value of your "a" exponent value "x" "a" exponent value "y" they add up result for the a ^ x + y is this more understandable?

  82. anonymous
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    Yes

  83. anonymous
    • 5 years ago
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    so can we try that again how to get x^3 from x^2

  84. anonymous
    • 5 years ago
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    (x)(x+x)?

  85. anonymous
    • 5 years ago
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    i just want it from x^2

  86. anonymous
    • 5 years ago
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    I'm sorry I have no idea what to do :(

  87. anonymous
    • 5 years ago
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    since it is already squared right (x^2)(x)=x how many more "x" do we need for x ^3

  88. anonymous
    • 5 years ago
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    x^2

  89. anonymous
    • 5 years ago
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    you have x^2 already

  90. anonymous
    • 5 years ago
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    but x^3 is made up of x^3 theres not anything left to do

  91. anonymous
    • 5 years ago
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    remember (x^2)(x)=x? ^ | value of exponent you are adding to the result

  92. anonymous
    • 5 years ago
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    lol if there is nothing then say nothing

  93. anonymous
    • 5 years ago
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    be more sure of you self now just give the reason

  94. anonymous
    • 5 years ago
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    nm that its probably more difficult to explain

  95. anonymous
    • 5 years ago
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    but do you understand how it works

  96. anonymous
    • 5 years ago
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    thats all i want to know

  97. anonymous
    • 5 years ago
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    How about we start over with a new problem, m^3+216n^3 Show me in steps how to solve this and ill try another on my own, this might work

  98. anonymous
    • 5 years ago
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    okeedokee

  99. anonymous
    • 5 years ago
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    lol this is differnt now because it has 2 variables in it

  100. anonymous
    • 5 years ago
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    okie now we can use something called difference of squares a^3+b^3= (a+b) (a^2-ab+b^2)

  101. anonymous
    • 5 years ago
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    ok

  102. anonymous
    • 5 years ago
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    first of all you must clearly understand the laws of exponents (or at least the, 1 i showed you, you will have to use it for these problems )

  103. anonymous
    • 5 years ago
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    we can go over it again i dont mind

  104. anonymous
    • 5 years ago
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    sure :)

  105. anonymous
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    okie i will put a sample problem (x^3) (x^3 )=x ^6 the exponents values of "x " add up together (d ^100) ( d ^ 90) = d ^ 190 exponents values of "d" add up together

  106. anonymous
    • 5 years ago
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    does this make sense :)

  107. anonymous
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    yes

  108. anonymous
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    "for now" <--------------------note i can go over with the rest of the laws but i hhave limited time this is what you will need difference of squares a^3+b^3= (a+b) (a^2-ab+b^2

  109. anonymous
    • 5 years ago
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    ok

  110. anonymous
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    m^3+216n^3 know what you are working with m^3 + 216n^3 ^ ^ first term | second term

  111. anonymous
    • 5 years ago
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    a^3+b^3= ( a+b ) (a^2- ab+ b^2) ^ ^ you only have "items" that you are going to be multiplying using FOIL with

  112. anonymous
    • 5 years ago
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    a^3+b^3= (a+b) (a^2-ab+b^2) | ^ ^ | this part wants the numbers squared to | be multiplied with L--------------------|

  113. anonymous
    • 5 years ago
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    how do we get a squared variable to be cubed? :)

  114. anonymous
    • 5 years ago
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    ^3?

  115. anonymous
    • 5 years ago
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    m^2 to m^3

  116. anonymous
    • 5 years ago
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    add another m?

  117. anonymous
    • 5 years ago
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    type how you would do that plz :)

  118. anonymous
    • 5 years ago
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    its correct but how would it look

  119. anonymous
    • 5 years ago
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    mathematically

  120. anonymous
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    give you a hint its not x+x

  121. anonymous
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    that is called adding like terms

  122. anonymous
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    it becomes 2x

  123. anonymous
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    remember this model (a ^ x) ( a ^y ) = a ^x+y

  124. anonymous
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    x is going to be your a variable

  125. anonymous
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    the values of your exponent to the x are?

  126. anonymous
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    i gave you already x^2 you have x^2 how much you need to get to x^3

  127. anonymous
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    (x ^ 2) ( x ^y ) = x ^2+y what will you need to get 3

  128. anonymous
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    as the exponent result

  129. anonymous
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    x

  130. anonymous
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    correct :)

  131. anonymous
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    so the value of that 1 that you are missing is just x

  132. anonymous
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    ^^

  133. anonymous
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    it is because x*x*x = x^3 but i think you know that

  134. anonymous
    • 5 years ago
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    do you know how to simplify radicals

  135. anonymous
    • 5 years ago
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    I do

  136. anonymous
    • 5 years ago
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    im sorry i have chat lag :(

  137. anonymous
    • 5 years ago
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    cube root of 216 is 6 but you want 6^2 and n^2 so you can do the FOIL

  138. anonymous
    • 5 years ago
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    sorries i gtg now :(

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