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mmbuckaroos

  • 2 years ago

Use the Factor Theorem to determine a polynomial equation, of lowest degree, that has only the indicated roots: 0 is a root of multiplicity 5, 2 is a double root

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  1. Mertsj
    • 2 years ago
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    x^5(x-2)^2=0

  2. satellite73
    • 2 years ago
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    ok i cant read. start with \[x^5(x-2)^2\] and multiply out

  3. mmbuckaroos
    • 2 years ago
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    so I would get \[x ^{5}+x ^{2}-4x+4\]

  4. satellite73
    • 2 years ago
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    no i don't think so

  5. satellite73
    • 2 years ago
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    \[x^5(x-2)^2=x^5(x-2)(x-2)=...\]

  6. mmbuckaroos
    • 2 years ago
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    where does the extra (x-2)^2 come from then that you have after the x^5?

  7. mmbuckaroos
    • 2 years ago
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    oh nvm.... so what am I multiplying out then? the (x-2)^2 or?

  8. Mertsj
    • 2 years ago
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    \[x^5(x-)(x-2)=x^5(x^2-4x+4)=x^7-4x^6+4x^5\]

  9. mmbuckaroos
    • 2 years ago
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    Oh ok so I just needed to continue multiplying to completly get rid of the parenthisis cool thank you... so what if it is the same question but it says 1/3 is a double root and -2 is a double root? do I set it up like this: f(x)=(x-1/3)^2(x+2)^2

  10. Mertsj
    • 2 years ago
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    Yes

  11. mmbuckaroos
    • 2 years ago
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    awesome thanks so much for the help!

  12. Mertsj
    • 2 years ago
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    yw

  13. mmbuckaroos
    • 2 years ago
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    Uh so I actually have one more silly question how do I get rid of the fraction to make that problem easier?

  14. mmbuckaroos
    • 2 years ago
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    NVM I figured it out:) thanks

  15. satellite73
    • 2 years ago
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    i wouldn't

  16. satellite73
    • 2 years ago
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    oh looks like you got it right? start with \((3x-1)^2\)

  17. mmbuckaroos
    • 2 years ago
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    Thanks:)

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