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anonymous

  • 5 years ago

2y(y^2-3y-2)-3(5y+2y^2-y^3)

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  1. sandra
    • 5 years ago
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    ok so you need to distribute out each term

  2. sandra
    • 5 years ago
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    so 1. 2y(y^2-3y-2) = (2y*y^2) - (2y *3y) - (2y*2) = 2y^3 -6y^2 -4y

  3. amistre64
    • 5 years ago
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    id "factor" out each term; but feel free to distribute them out :)

  4. sandra
    • 5 years ago
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    hmmm, good point

  5. anonymous
    • 5 years ago
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    thank u!

  6. sandra
    • 5 years ago
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    well I'll do it this way, then you sow that way =)

  7. sandra
    • 5 years ago
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    2. 3(5y+2y^2-y^3) = (3*5y) + (3*2y^2) - (3*y^3) = 15y + 6y^2 - 3y^3

  8. amistre64
    • 5 years ago
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    youre doing fine :)

  9. sandra
    • 5 years ago
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    now subtract them: 2y^3 -6y^2 -4y - (15y + 6y^2 - 3y^3) = 2y^3 -6y^2 -4y -15y -6y^2 +3y^3

  10. sandra
    • 5 years ago
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    and combine like terms (all of the y's, y^2 and y^3): -y^3 -12y^2 -19y

  11. sandra
    • 5 years ago
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    oops one second!

  12. sandra
    • 5 years ago
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    first one is 5y^3 heh, since 2+3=5, not -1 =)

  13. sandra
    • 5 years ago
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    so I get 5y^3 -12y^2 -19y

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