Question

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

Answer 1

Do you know about the Rational Root Theorem?

Answer 2

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

Answer 3

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

Answer 4

Don't forget the +/- sign

Answer 5

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

Answer 6

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

Answer 7

Except this time, your leading coefficient (for 4x³) isn't 1, so the number of possible rational roots varies even more :)

Answer 8

Lots of synthetic division to do :P you could prob make use of the upper/lower bounds, but.. it may not save much time.

Answer 9

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

Answer 10

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

Answer 11

so then, the possible rational zeroes are \[\pm1, \pm2, \pm3, \pm6\] and \[\pm1, \pm 2, \pm4\] right?

Answer 12

They can't all be integers, that's too good to be true o.O

Answer 13

Not quite, you have to divide the factors... \[x \pm \frac{ p }{q }\]

Answer 14

\[x = \pm \frac{ p }{q } \]

Answer 15

Oh! So now I divide the factors of 4 by the factors of 6, right?

Answer 16

\[p = \pm1, \pm2, \pm3, \pm6 \] \[q = \pm1, \pm 2, \pm4\]

Answer 17

Factors of 6 / factors of 4 (one at a time, of course)

Answer 18

oh, so i divide the factors of 6 by the factors of 4

Answer 19

BINGO!

Answer 20

\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 } ,\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 2 }{ 2 },\frac{ 2 }{ 4 },\frac{ 3 }{ 1 }, \frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 2 }, and \frac{ 6 }{ 4 }\]

Answer 21

right?

Answer 22

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.

Answer 23

Wow!

Answer 24

\[\frac{ 1 }{ 1 }+\frac{ 2 }{ 1 }+\frac{ 3 }{ 1 }+\frac{ 6 }{ 1 }=\frac{ 12 }{ 1 }\]

Answer 25

wait, do I combine and reduce, or just reduce?

Answer 26

Why add?

Answer 27

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.

Answer 28

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

Answer 29

ok

Answer 30

\[\pm\frac{ 1 }{ 1 },\pm\frac{ 1 }{ 2 },\pm\frac{ 1 }{ 4 },\pm\frac{ 2 }{ 1 },\pm\frac{ 2 }{ 1 },(\pm\frac{ 1 }{ 1 }),(\pm\frac{ 1 }{ 2 }),\pm\frac{ 3 }{ 1 },\pm\frac{ 3 }{ 2 },\pm\frac{ 3 }{ 4 },\pm\frac{ 6 }{ 1 },(\pm\frac{ 3 }{ 1 }),\pm\frac{ 6 }{ 4 }\]

Answer 31

right?

Answer 32

Remove the repeated ones

Answer 33

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

Answer 34

oh, so 6/4 should be 3/2

Answer 35

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

Answer 36

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

Answer 37

\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }\]

Answer 38

As we said, \(\frac{\rm stuff}{1} = \rm stuff\)

Answer 39

so then the whole zeroes are 1, 2, 3, and 6, the factors of 6

Answer 40

That's right

Answer 41

Y u no write 3/2 instead of 6/4?
:D

Answer 42

Y u no write 1 1/2 or 1.5 instead of 3/2?
:D

Answer 43

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

Answer 44

so , the remaining fractions are 1/2, 1/4, 3/2, and 3/4

Answer 45

\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }\]= ± 1, 1/2, 1/4, 2, 3, 3/2, 3/4, 6\

note the ± needs to be in front of each one

Answer 46

ok

Answer 47

so, are the remaining fractions possible rational zeroes?

Answer 48

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