anonymous
  • anonymous
Find any rational roots of P(x). P(x)=3x^4+2x^3-9x^2+4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
try 1, -1 first then 2, -2 then 4,-4 if these don't work we have to move on to fractions
anonymous
  • anonymous
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
anonymous
  • anonymous
So what exactly is the textbook way for your textbook

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More answers

anonymous
  • anonymous
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
anonymous
  • anonymous
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"
anonymous
  • anonymous
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.
anonymous
  • anonymous
ya thats the way i am just having trouble relating it to this specifc problem
anonymous
  • anonymous
1 works by your eyeballs because if \[P(x)=3x^4+2x^3-9x^2+4\] then \[P(1)=3+2-9+4=0\]
anonymous
  • anonymous
well for this one i would guess -2
anonymous
  • anonymous
but its just a guess
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \((x-1)(x-1)(something)\)
anonymous
  • anonymous
x+2
anonymous
  • anonymous
but there really isn't an exact method to it is all.

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