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DeadShot
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What are the possible rational zeros of f(x) = x4 + 6x3  3x2 + 17x  15?
 one year ago
 one year ago
DeadShot Group Title
What are the possible rational zeros of f(x) = x4 + 6x3  3x2 + 17x  15?
 one year ago
 one year ago

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terenzreignz Group TitleBest ResponseYou've already chosen the best response.2
The rational root test... Given a polynomial with integer coefficients \[\large a_nx^n + a_{n1}x^{n1}....+a_1x+a_0\] If this polynomial is to have a rational root (zero), then it will ALWAYS be of the form \[\huge \pm \frac{p}{q}\] \[\large where \ p \ is \ a \ factor \ of \ a_0\]\[\large and \ q \ is \ a \ factor \ of \ a_n\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.2
Lucky for you, it seems \[\huge a_n = 1\] This simplifies things...
 one year ago

DeadShot Group TitleBest ResponseYou've already chosen the best response.0
so, if \[a _{n}=1\] ten how do i solve for \[a _{0}\] ?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.2
You don't *solve* for \[\large a_0\]it's given. It's the last term in the polynomial, the constant, 15. What are the factors of 15?
 one year ago

DeadShot Group TitleBest ResponseYou've already chosen the best response.0
1, 3, 5, and 15, right?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.2
That's right. So those are the numerators if ever you're to have a rational root. But your leading coefficient is 1, so as I said, that simplifies things. What are your possible rational roots, then?
 one year ago

DeadShot Group TitleBest ResponseYou've already chosen the best response.0
\[\pm1, \pm3, \pm5, and \pm15\] right?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.2
Bingo. :)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.2
No problem
 one year ago
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