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let me see if I can still remember this... you need all factors of the last digit which is 50 and then test it to see if the result comes to 0 then use long division to grab the other roots awwwwwww crud.. ok so we need to use synthetic division and make sure that our remainder is 0 ???
yeah if th.......... man........ I like the other way though... like find all factors, test it out in the function to see if we have 0 and then use long division on it, but I guess we can use synthetic to make the process go faster.
I remember that you have to do synthetic division and I know how to do that, but I don't know how to find the zeros. Was it something to do with the exponents and putting them in order and then taking them and testing them all out individually?
find all possible factors of the number 50 which is 1,2,5,10,25,50 but we have to consider the negative versions too so -1,-2,-5,-10,-25,-50
12 numbers to test out .......
Aren't some of those numbers ones you would be using for the table itself? So at the top it would be something like... 1 2 −25 50 then you would test the numbers with those?
I'm getting -5, 5, and 2, but those are probably waaay off.
hmmm I've done this differently... once I got all of my possible factors, I plug them back into the equation and see if I end up with a 0. If I do, it's a root. If not, toss that number out and try again. ANd then I use long division (because I was taught how to do this differently and prefer that style) did you plug in x = -5,5, and 2 into the equation?
I tested these and as long as I didn't screw up my long division these should be the roots. I'm not completely sure though. :/
either way ... it's right.
but I would've done long division... even though it takes forever.
you can check @mintalicarth by finding the following product (x-5)(x+5)(x-2)
|dw:1438755053787:dw| so when I plugged in x = 2 for the original equation, I got a 0 , so that's one root . we use long division or synthetic division to find the rest of the roots
hmmm it's been a while since I've done synthetic... maybe I should do that :P |dw:1438755208744:dw|
so if we use difference of squares formula |dw:1438755285695:dw| solving for x gives x = 5, and -5
so we found one root by testing it into the original equation and found the remaining roots using long division (or synthetic division) so.. x = 5,-5,2
So if that's the solution set, would that be all the rational roots that are possible, or should I include the ones you said earlier?
solution set meaning which x value works...
x = 5,-5,2
Basically I'm just making sure I have everything: a. -1, -2, -5, -10, -25, -50, 1, 2, 5, 10, 25, 50 b. 2 c. -5,5,2
Thanks for all the help. :3