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

1. ParthKohli

Do you know about the Rational Root Theorem?

2. agent0smith

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

4. agent0smith

Don't forget the +/- sign

5. terenzreignz

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

6. agent0smith

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

7. terenzreignz

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

8. agent0smith

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

9. terenzreignz

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

10. agent0smith

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?

12. terenzreignz

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

13. agent0smith

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

14. agent0smith

$x = \pm \frac{ p }{q }$

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

16. agent0smith

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

17. agent0smith

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

19. terenzreignz

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?

22. terenzreignz

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.

23. ParthKohli

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?

26. ParthKohli

27. terenzreignz

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.

28. terenzreignz

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?

32. agent0smith

Remove the repeated ones

33. terenzreignz

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

35. ParthKohli

$\pm \dfrac{1}{1} = \pm 1$:-)

36. agent0smith

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

$\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }$

38. ParthKohli

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

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

40. ParthKohli

That's right

41. terenzreignz

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

42. ParthKohli

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

43. terenzreignz

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

45. agent0smith

$\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }$= ± 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?