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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
 one year 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
 one year ago

Freyes Group TitleBest ResponseYou've already chosen the best response.0
So what exactly is the textbook way for your textbook
 one year 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
 one year 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"
 one year 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.
 one year 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
 one year 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 \[P(1)=3+29+4=0\]
 one year ago

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

Freyes Group TitleBest ResponseYou've already chosen the best response.0
but its just a guess
 one year 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
 one year 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
 one year 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 \((x1)(x1)(something)\)
 one year ago

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