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anonymous

  • 4 years ago

What are the possible rational zeros of f(x) = 2x3 – 4x2 – 7x + 14 ?

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  1. y2o2
    • 4 years ago
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    2x3 – 4x2 – 7x + 14 = 2x²(x-2)-7(x-2) = (x-2)(2x²-7) = 0 you can proceed...

  2. nenadmatematika
    • 4 years ago
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    just 2

  3. anonymous
    • 4 years ago
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    you can also use the rational root theorem..

  4. nenadmatematika
    • 4 years ago
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    the other zeros are +-sqrt(7/2) but they're not ratonal numbers

  5. anonymous
    • 4 years ago
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    possible rational roots would include all integer factors of 14 divided by all integer factors of the leading coefficient.

  6. anonymous
    • 4 years ago
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    so +-14/2,+-7/2,+-7/1,+-7/1

  7. anonymous
    • 4 years ago
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    that doesn't mean all are roots, it just means these are all of the possible roots.

  8. anonymous
    • 4 years ago
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    oops I forgot a couple ....+-2/2, +-2/1

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