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Math Help - Factor *!URGENT!*

  1. #1
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    Factor *!URGENT!*

    i need help with this problem

    Factor:
    (3ax^2+2b^2y+axy)^2
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  2. #2
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    Quote Originally Posted by AHDDM
    i need help with this problem

    Factor:
    (3ax^2+2b^2y+axy)^2
    Not sure what you want.
    [ax(3x+y)+2b^2y]^2
    (ax)^2(3x+y)^2+4ab^2xy(3x+y)+4b^4y^2
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  3. #3
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    Hello, AHDDM!

    Are you sure the instructions say "factor"?


    Factor: . (3ax^2 + 2b^2y + axy)^2

    Simple!

    (3ax^2 + 2b^2y + axy)^2\;=\;(3ax^2 + 3b^2y + axy) \cdot(3ax^2 + 2b^2y + axy)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    It's more likely that they want us to multiply . . .

    After a lot of multiplying and combining, we get:
    . . 9a^2x^4 + 12ab^2x^2y + 6a^2x^3y + 4b^4y^2 + 4ab^2xy^2 + a^2x^2y^2

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  4. #4
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    i got wat soroban got after multipling so ima go with tat.

    can you guys also help me with this

    (2x-7)^6
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  5. #5
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by AHDDM
    (2x-7)^6
    what's the question, that number is already factored...
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  6. #6
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    if its already factored then im ok thats what i got from 64x^6-y^6
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  7. #7
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    Quote Originally Posted by AHDDM
    if its already factored then im ok thats what i got from 64x^6-y^6
    No that is wrong.
    You need to use the binomial theorem.
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  8. #8
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Quick
    what's the question, that number is already factored...
    I'm sorry, let me rephrase this to be:
    Quote Originally Posted by Quick
    what's the question, that number is already as simple as it will get...
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