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mmbuckaroos Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Mertsj Group Title
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    x^5(x-2)^2=0

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

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

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

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

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

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

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

    • 2 years ago
  9. mmbuckaroos Group Title
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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

    • 2 years ago
  10. Mertsj Group Title
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    Yes

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

    • 2 years ago
  12. Mertsj Group Title
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    yw

    • 2 years ago
  13. mmbuckaroos Group Title
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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?

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

    • 2 years ago
  15. satellite73 Group Title
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    i wouldn't

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

    • 2 years ago
  17. mmbuckaroos Group Title
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    Thanks:)

    • 2 years ago
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