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Find any rational roots of P(x). P(x)=3x^4+2x^3-9x^2+4


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try 1, -1 first then 2, -2 then 4,-4 if these don't work we have to move on to fractions
there 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
So what exactly is the textbook way for your textbook

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there is no other way to do it you have to check. if that is "trial and error' that is what you need to do
Well, in practicality my teacher told me to graph the function.. and then you can use synthetic division for the ones that aren't "exact"
But 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.
ya thats the way i am just having trouble relating it to this specifc problem
1 works by your eyeballs because if \[P(x)=3x^4+2x^3-9x^2+4\] then \[P(1)=3+2-9+4=0\]
well for this one i would guess -2
but its just a guess
add up the coefficients and see that you get 0 then you can factor as \[(x-1)\times (something)\] and then you can either factor the "something" or else use the quadratic formula
you only need to find one zero, then factor and end up with a quadratic, and you can always find the zeros of any quadratic
scratch 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 \((x-1)(x-1)(something)\)
but there really isn't an exact method to it is all.

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