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

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

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

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

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

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

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

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

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

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

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

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

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

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

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
BINGO!
 one year ago

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Why add?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

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

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

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

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

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
That's right
 one year ago

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

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

agent0smith Group TitleBest 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
 one year ago

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

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