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How do I find all possible rational zeros of the polynomial function f(x) = 4x^3 - 5x^2 + 9x - 6?

Mathematics
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Do you know about the Rational Root Theorem?
Use the rational zero theorem \[x = \frac{ p }{ q }\] p is a factor of the constant term (-6 in this case) and q is a factor of the coefficient on the highest power (4 here). So find factors of 6 and factors of 4. Then use synthetic division to check
okay, so then it would be 1, 2, 3, and 6, and then 1, 2, and 4

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Other answers:

Don't forget the +/- sign
Hey, @DeadShot This seems familiar :D You can't just answer a question and then forget the concepts that led you to the answer, they might be useful in a future question! :) http://openstudy.com/study#/updates/511f7c31e4b06821731c6be5
heh, you might want to read over that ^^^ question again, @DeadShot
Except this time, your leading coefficient (for 4x³) isn't 1, so the number of possible rational roots varies even more :)
Lots of synthetic division to do :P you could prob make use of the upper/lower bounds, but.. it may not save much time.
No need for dividing, the question only asked for the "possible" rational zeros... I don't really want to exert more effort than necessary XD
Oh i thought it asked for actual zeroes... you're right.
so then, the possible rational zeroes are \[\pm1, \pm2, \pm3, \pm6\] and \[\pm1, \pm 2, \pm4\] right?
They can't all be integers, that's too good to be true o.O
Not quite, you have to divide the factors... \[x \pm \frac{ p }{q }\]
\[x = \pm \frac{ p }{q } \]
Oh! So now I divide the factors of 4 by the factors of 6, right?
\[p = \pm1, \pm2, \pm3, \pm6 \] \[q = \pm1, \pm 2, \pm4\]
Factors of 6 / factors of 4 (one at a time, of course)
oh, so i divide the factors of 6 by the factors of 4
BINGO!
\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 } ,\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 2 }{ 2 },\frac{ 2 }{ 4 },\frac{ 3 }{ 1 }, \frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 2 }, and \frac{ 6 }{ 4 }\]
right?
Wow, so much effort. Don't forget to reduce them to lowest terms, though. Did you notice that 6/4 = 3/2 ? :P Also, don't forget the +/- signs. Either the positives or the negatives of these may be rational roots.
Wow!
\[\frac{ 1 }{ 1 }+\frac{ 2 }{ 1 }+\frac{ 3 }{ 1 }+\frac{ 6 }{ 1 }=\frac{ 12 }{ 1 }\]
wait, do I combine and reduce, or just reduce?
Why add?
Just reduce them to lowest terms. The list you provided: \[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 } ,\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 2 }{ 2 },\frac{ 2 }{ 4 },\frac{ 3 }{ 1 }, \frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 2 }, and \frac{ 6 }{ 4 }\] is correct, but just reduce them to lowest terms, and notice which ones are equal.
Also, don't forget the \[\huge \pm\] sign.
ok
\[\pm\frac{ 1 }{ 1 },\pm\frac{ 1 }{ 2 },\pm\frac{ 1 }{ 4 },\pm\frac{ 2 }{ 1 },\pm\frac{ 2 }{ 1 },(\pm\frac{ 1 }{ 1 }),(\pm\frac{ 1 }{ 2 }),\pm\frac{ 3 }{ 1 },\pm\frac{ 3 }{ 2 },\pm\frac{ 3 }{ 4 },\pm\frac{ 6 }{ 1 },(\pm\frac{ 3 }{ 1 }),\pm\frac{ 6 }{ 4 }\]
right?
Remove the repeated ones
You missed a spot :) And anyway, the point of me asking you to reduce to lowest terms is... that^ Remove the repeated ones. Again, you missed a spot, 6/4 is not yet in lowest terms :D
oh, so 6/4 should be 3/2
\[\pm \dfrac{1}{1} = \pm 1\]:-)
And 2/1, 3/1, 6/1...
\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }\]
As we said, \(\frac{\rm stuff}{1} = \rm stuff\)
so then the whole zeroes are 1, 2, 3, and 6, the factors of 6
That's right
Y u no write 3/2 instead of 6/4? :D
Y u no write 1 1/2 or 1.5 instead of 3/2? :D
Well, the thing is, 3/2 is there, and so is 6/4. This town ain't big enough for the two of them :D
so , the remaining fractions are 1/2, 1/4, 3/2, and 3/4
\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }\]= ± 1, 1/2, 1/4, 2, 3, 3/2, 3/4, 6\ note the ± needs to be in front of each one
ok
so, are the remaining fractions possible rational zeroes?
Oh, I get this one now! Thanks for the help!

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