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

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Freyes Group TitleBest ResponseYou've already chosen the best response.0
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"
 2 years ago

Freyes Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

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

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

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

Freyes Group TitleBest ResponseYou've already chosen the best response.0
but its just a guess
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
add 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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 \((x1)(x1)(something)\)
 2 years ago

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