a function is shown below: f(x)=x^3+4x^2-x-4 part a.) what are the factors of f(x) part b.) what are the zeros of f(x) part c.) what are the steps you would follow to graph f(x) describe the end behavior of the graph.
to factor this, you can factor by grouping like x^3 + 4x^2 have x^2 in common so you can factor that part into x^2 ( x + 4) and same idea with -x-4 part they have a negative in common so you can get - (x+4) then x^2(x+4) and -(x+4) have (x+4) in common so you can factor that out and get (x+4)(x^2 -1) then know that x^2 - 1 can be factored further into (x+1)(x-1)
that's part a asnswer @jigglypuff314
I did not give the actual answer, I hinted toward how you could get the answer from (x+4)(x^2 - 1) x^2-1 can be factored further into (x+1)(x-1)
im really confussed right now i have been trying to answer this question for the last hour and a half, i need to get this done like now and everybody who tried to help me is just confusing me more and now i'm starting to get frustrated my brain literally isn't even function right now
I'm sorry, I was trying to say that you can plug in (x+1)(x-1) for (x^2-1) that was in (x+4)(x^2-1) if you plug in like that, what would you get? :)
its not your fault just having a long day... and i guess it would be (x+4)(X+1)(x-1)?
correct!!! :) so that's for part A
then to find the zeros (x+4)(x+1)(x-1) = 0 you can get three numbers x+4=0 x+1=0 x-1=0 solve for x for each :)
x is 0
mmm not quite what I meant :) like for x+4 = 0 subtract both sides by 4 -4 -4 so x = -4 would be one of your answers :) there are two more x+1 = 0 x - 1 = 0 solve for x for both :)
-1 and +1?
Correct!!! :D great job! so for part B you've got -4, -1, and +1 :)
:) i really appreciate your help, what about part c?
well you can put the points on the x-axis x = -4, -1, and +1 then to test where to draw the curve when x = 0 (0+4)(0+1)(0-1) = -4 so you'll get something like |dw:1395011959727:dw|
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