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andriod09

  • 2 years ago

HELP PLEASE! EQUATION IS IN COMMENTS

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  1. andriod09
    • 2 years ago
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    x^2y-5xy+6y+7x^2-35x+42

  2. Nicole<3Algebra
    • 2 years ago
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    solving for x?

  3. andriod09
    • 2 years ago
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    factoring

  4. jasonxx
    • 2 years ago
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    xy(x-5)+6y+7x(x-5)+42=0

  5. jasonxx
    • 2 years ago
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    (xy+7x)(x-5)+(6y+42)=0 x(y+7)(x-5)+6(y+7)=0 {x(x-5)+6}(y+7)=0....i am not sure tho

  6. jasonxx
    • 2 years ago
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    @andriod09 does that help?

  7. andriod09
    • 2 years ago
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    no, not really. but i have the answer.

  8. dhiab
    • 2 years ago
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    hello solving for x x(x-5)+6=0 x^2-5x+6=0 delta =1 x1=3 and x2=3

  9. dhiab
    • 2 years ago
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    x2 =2

  10. andriod09
    • 2 years ago
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    I AM NOT SOLVING FOR "X"!!! I AM FACTORING THE POLYNOMIAL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  11. Nicole<3Algebra
    • 2 years ago
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    ^^^^got so mad

  12. dhiab
    • 2 years ago
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    your question is "HELP PLEASE! EQUATION IS IN COMMENTS"

  13. andriod09
    • 2 years ago
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    "andriod09 0 factoring 10 minutes agoDelete"

  14. jasonxx
    • 2 years ago
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    is it incomplete?

  15. andriod09
    • 2 years ago
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    all i had to do is factor the polynomial. thats it. i already have the answer.

  16. jasonxx
    • 2 years ago
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    that's it

  17. andriod09
    • 2 years ago
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    but its not the correct answer

  18. jasonxx
    • 2 years ago
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    then do it other way

  19. andriod09
    • 2 years ago
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    the correct answer is: (x-2)(x-3)(y+7)

  20. jasonxx
    • 2 years ago
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    dude factor my last equation...you will get this

  21. jasonxx
    • 2 years ago
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    (x^2-5x+6)(y+7)=0 (x-2)(x-3)(y+7)=0

  22. andriod09
    • 2 years ago
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    no, your last equation as a quadratic equation, this is just a normal polynomial

  23. jasonxx
    • 2 years ago
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    {x(x-5)+6}(y+7)=0 {x^2-5x+6}(y+7)=0 (x^2-3x-2x+6)(y+7)=0 (x-2)(x-3)(y+7)=0 @Nicole<3Algebra can you help further

  24. jasonxx
    • 2 years ago
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    @andriod09 hope it is clear to you now?

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