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Do you know about the Rational Root Theorem?

okay, so then it would be 1, 2, 3, and 6, and then 1, 2, and 4

Don't forget the +/- sign

heh, you might want to read over that ^^^ question again, @DeadShot

Oh i thought it asked for actual zeroes... you're 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!

right?

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?

Also, don't forget the
\[\huge \pm\]
sign.

ok

right?

Remove the repeated ones

oh, so 6/4 should be 3/2

\[\pm \dfrac{1}{1} = \pm 1\]:-)

And 2/1, 3/1, 6/1...

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

ok

so, are the remaining fractions possible rational zeroes?

Oh, I get this one now! Thanks for the help!