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anonymous
 3 years ago
What are the possible rational zeros of f(x) = x4 + 6x3  3x2 + 17x  15?
anonymous
 3 years ago
What are the possible rational zeros of f(x) = x4 + 6x3  3x2 + 17x  15?

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terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2The 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 \[\huge \pm \frac{p}{q}\] \[\large where \ p \ is \ a \ factor \ of \ a_0\]\[\large and \ q \ is \ a \ factor \ of \ a_n\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so, if \[a _{n}=1\] ten how do i solve for \[a _{0}\] ?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2You 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01, 3, 5, and 15, right?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2That'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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\pm1, \pm3, \pm5, and \pm15\] right?
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