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InsanelyChaotic

  • 3 years ago

2x3 - 5x2y + xy2 + 2y3 a. Simplify (2x2 + y2)(x - 2y) = ? b. Simplify (2x + y)(x2 -3xy + 2y 2 ) = ?

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  1. timo86m
    • 3 years ago
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    where is c and d?

  2. timo86m
    • 3 years ago
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    \[2\,{x}^{3}-5\,{x}^{2}y+x{y}^{2}+2\,{y}^{3}\] did you mean that?

  3. InsanelyChaotic
    • 3 years ago
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    yes

  4. timo86m
    • 3 years ago
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    ooh where are c and d then :) i got my own answer can i see c and d?

  5. InsanelyChaotic
    • 3 years ago
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    theyre not right for this

  6. timo86m
    • 3 years ago
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    so far i think it is b but i feel it can be simplified further :)

  7. InsanelyChaotic
    • 3 years ago
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    elimination

  8. InsanelyChaotic
    • 3 years ago
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    its a

  9. InsanelyChaotic
    • 3 years ago
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    ;)

  10. timo86m
    • 3 years ago
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    was it a?

  11. timo86m
    • 3 years ago
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    well i get :) (-y+x)*(x-2*y)*(y+2*x) fully factored

  12. InsanelyChaotic
    • 3 years ago
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    idk

  13. timo86m
    • 3 years ago
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    :) i'll check a and b :)

  14. timo86m
    • 3 years ago
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    i dont think it is a :(

  15. timo86m
    • 3 years ago
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    now for b

  16. InsanelyChaotic
    • 3 years ago
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    its b!!!!!!!!!!!!!!!!!!:)))))))))))))))))))))))))))

  17. timo86m
    • 3 years ago
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    told ya ;)

  18. timo86m
    • 3 years ago
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    and i check to and it is b :D

  19. timo86m
    • 3 years ago
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    If you know algrabeic long division then there is a theory of if you divide your given polynomial in this case 2x3 - 5x2y + xy2 + 2y3 by a suspected binomial or trinomial (2x + y) or (x2 -3xy + 2y 2 ) ^ those are from B the 2 expressions inside the ( ) and you get 0 then that trinomial or binomial is a factor and therfore part of the answer :)

  20. timo86m
    • 3 years ago
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    if you get a remainder of 0 :P

  21. timo86m
    • 3 years ago
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    in other words 2x3 - 5x2y + xy2 + 2y3 -------------------- = (x2 -3xy + 2y 2 ) with no remainder (2x + y) and obviously 2x3 - 5x2y + xy2 + 2y3 --------------------- = 2x+y with no remainder (x2 -3xy + 2y 2 )

  22. timo86m
    • 3 years ago
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    But if you dont know algabreic long division dont bother with that method :P

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