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mskyeg
 3 years ago
Find any rational roots of P(x).
P(x)=3x^4+2x^39x^2+4
mskyeg
 3 years ago
Find any rational roots of P(x). P(x)=3x^4+2x^39x^2+4

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satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0try 1, 1 first then 2, 2 then 4,4 if these don't work we have to move on to fractions

mskyeg
 3 years ago
Best ResponseYou've already chosen the best response.0there is more than 1 so even if they work there wtill might be a fraction. Oh and with this i have to show my work and do it the textbook way not by trial and error

Freyes
 3 years ago
Best ResponseYou've already chosen the best response.0So what exactly is the textbook way for your textbook

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0there is no other way to do it you have to check. if that is "trial and error' that is what you need to do

Freyes
 3 years ago
Best ResponseYou've already chosen the best response.0Well, in practicality my teacher told me to graph the function.. and then you can use synthetic division for the ones that aren't "exact"

Freyes
 3 years ago
Best ResponseYou've already chosen the best response.0But unless you have some sort of sixth sense of math you really don't have a set method.. unless your textbook does, in which case i'd want to know.

mskyeg
 3 years ago
Best ResponseYou've already chosen the best response.0ya thats the way i am just having trouble relating it to this specifc problem

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.01 works by your eyeballs because if \[P(x)=3x^4+2x^39x^2+4\] then \[P(1)=3+29+4=0\]

Freyes
 3 years ago
Best ResponseYou've already chosen the best response.0well for this one i would guess 2

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0add up the coefficients and see that you get 0 then you can factor as \[(x1)\times (something)\] and then you can either factor the "something" or else use the quadratic formula

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0you only need to find one zero, then factor and end up with a quadratic, and you can always find the zeros of any quadratic

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0scratch that, this is a polynomial of degree 4, so you factor and end up with a cubic when you do that, you will see that 1 is a zero of the cubic polynomial as well, so you will factor again as \((x1)(x1)(something)\)

Freyes
 3 years ago
Best ResponseYou've already chosen the best response.0but there really isn't an exact method to it is all.
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