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Keandrap15

  • 2 years ago

What does 13m2n3p + 21mp2 equal when completely factored? (A)13mp(mn3 + 2p) (B)m2p2(13n3 + 21) (C)mp(13mn3 + 21p) (D)mnp(13mn2 + 21p)

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  1. AKANicole
    • 2 years ago
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    i know this one!

  2. abb0t
    • 2 years ago
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    It looks like you can factor out an m and p.

  3. AKANicole
    • 2 years ago
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    let me do it!

  4. AKANicole
    • 2 years ago
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    dude!

  5. abb0t
    • 2 years ago
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    Also. combine all numbers since they are just constants. So multiply 13 x 2 x 3 :)

  6. AKANicole
    • 2 years ago
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    um nvm i dont no this

  7. abb0t
    • 2 years ago
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    Put constants and variables together but keep them separate so you can see

  8. AKANicole
    • 2 years ago
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    idk i think so

  9. AKANicole
    • 2 years ago
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    let meh try

  10. abb0t
    • 2 years ago
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    wait, nevermind, are those 3 and 2's mean they are raised to that power?

  11. AKANicole
    • 2 years ago
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    |dw:1365966144717:dw|

  12. AKANicole
    • 2 years ago
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    sry bad at this i am young ok

  13. Keandrap15
    • 2 years ago
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    abb0t yes the 2 and 3 are raised to the power

  14. abb0t
    • 2 years ago
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    \(13m^2n^3p +21m p^2\) is that correct?

  15. abb0t
    • 2 years ago
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    Ok, simply factor the lowest common factor between them :)

  16. Keandrap15
    • 2 years ago
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    Yes that is correct

  17. abb0t
    • 2 years ago
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    Your two common variables are "m" and "p" That's because you have both of them there. There's no "n" variable on the right side. Now, notice that you have ONE "p" on the left and ONE "m" on the left. Those are your common factors.

  18. abb0t
    • 2 years ago
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    There's also no two common factors between 13 and 21, so you can just leave them in there. In essence, you're taking out ONE "m" and ONE "p" only.

  19. abb0t
    • 2 years ago
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    Now, can you make what your answer is?

  20. Keandrap15
    • 2 years ago
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    would it be C?

  21. Nurali
    • 2 years ago
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    correct

  22. abb0t
    • 2 years ago
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    Yep.

  23. Keandrap15
    • 2 years ago
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    thank you

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