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What are the possible rational zeros of f(x) = 2x3 – 4x2 – 7x + 14 ?
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2x3 – 4x2 – 7x + 14 = 2x²(x-2)-7(x-2) = (x-2)(2x²-7) = 0 you can proceed...
just 2
you can also use the rational root theorem..
the other zeros are +-sqrt(7/2) but they're not ratonal numbers
possible rational roots would include all integer factors of 14 divided by all integer factors of the leading coefficient.
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so +-14/2,+-7/2,+-7/1,+-7/1
that doesn't mean all are roots, it just means these are all of the possible roots.
oops I forgot a couple ....+-2/2, +-2/1
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