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

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Do you know about the Rational Root Theorem?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3Use 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

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0okay, so then it would be 1, 2, 3, and 6, and then 1, 2, and 4

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3Don't forget the +/ sign

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Hey, @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

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3heh, you might want to read over that ^^^ question again, @DeadShot

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Except this time, your leading coefficient (for 4x³) isn't 1, so the number of possible rational roots varies even more :)

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3Lots of synthetic division to do :P you could prob make use of the upper/lower bounds, but.. it may not save much time.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1No need for dividing, the question only asked for the "possible" rational zeros... I don't really want to exert more effort than necessary XD

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3Oh i thought it asked for actual zeroes... you're right.

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0so then, the possible rational zeroes are \[\pm1, \pm2, \pm3, \pm6\] and \[\pm1, \pm 2, \pm4\] right?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1They can't all be integers, that's too good to be true o.O

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3Not quite, you have to divide the factors... \[x \pm \frac{ p }{q }\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3\[x = \pm \frac{ p }{q } \]

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0Oh! So now I divide the factors of 4 by the factors of 6, right?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3\[p = \pm1, \pm2, \pm3, \pm6 \] \[q = \pm1, \pm 2, \pm4\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3Factors of 6 / factors of 4 (one at a time, of course)

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0oh, so i divide the factors of 6 by the factors of 4

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 } ,\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 2 }{ 2 },\frac{ 2 }{ 4 },\frac{ 3 }{ 1 }, \frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 2 }, and \frac{ 6 }{ 4 }\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Wow, 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.

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 1 }+\frac{ 2 }{ 1 }+\frac{ 3 }{ 1 }+\frac{ 6 }{ 1 }=\frac{ 12 }{ 1 }\]

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0wait, do I combine and reduce, or just reduce?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Just 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.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Also, don't forget the \[\huge \pm\] sign.

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0\[\pm\frac{ 1 }{ 1 },\pm\frac{ 1 }{ 2 },\pm\frac{ 1 }{ 4 },\pm\frac{ 2 }{ 1 },\pm\frac{ 2 }{ 1 },(\pm\frac{ 1 }{ 1 }),(\pm\frac{ 1 }{ 2 }),\pm\frac{ 3 }{ 1 },\pm\frac{ 3 }{ 2 },\pm\frac{ 3 }{ 4 },\pm\frac{ 6 }{ 1 },(\pm\frac{ 3 }{ 1 }),\pm\frac{ 6 }{ 4 }\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3Remove the repeated ones

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1You 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

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0oh, so 6/4 should be 3/2

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\pm \dfrac{1}{1} = \pm 1\]:)

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3And 2/1, 3/1, 6/1...

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0As we said, \(\frac{\rm stuff}{1} = \rm stuff\)

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0so then the whole zeroes are 1, 2, 3, and 6, the factors of 6

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Y u no write 3/2 instead of 6/4? :D

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Y u no write 1 1/2 or 1.5 instead of 3/2? :D

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Well, the thing is, 3/2 is there, and so is 6/4. This town ain't big enough for the two of them :D

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0so , the remaining fractions are 1/2, 1/4, 3/2, and 3/4

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{ 1 }{ 1 },\frac{ 1 }{ 2 },\frac{ 1 }{ 4 },\frac{ 2 }{ 1 },\frac{ 3 }{ 1 },\frac{ 3 }{ 2 },\frac{ 3 }{ 4 },\frac{ 6 }{ 1 },\frac{ 6 }{ 4 }\]= ± 1, 1/2, 1/4, 2, 3, 3/2, 3/4, 6\ note the ± needs to be in front of each one

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0so, are the remaining fractions possible rational zeroes?

DeadShot
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I get this one now! Thanks for the help!
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