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- anonymous

Can someone help me out with Rational Zero Theorem problems?
x^3 - 2x^2 - 25 + 50 = 0
https://i.imgur.com/NLjwYRX.png

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- anonymous

- schrodinger

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- UsukiDoll

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 ???

- UsukiDoll

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.

- anonymous

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?

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- UsukiDoll

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

- UsukiDoll

12 numbers to test out .......

- anonymous

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?

- anonymous

I'm getting -5, 5, and 2, but those are probably waaay off.

- UsukiDoll

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?

- anonymous

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. :/

- UsukiDoll

either way ... it's right.

- UsukiDoll

but I would've done long division... even though it takes forever.

- freckles

you can check @mintalicarth
by finding the following product
(x-5)(x+5)(x-2)

- UsukiDoll

|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

- UsukiDoll

hmmm it's been a while since I've done synthetic... maybe I should do that :P
|dw:1438755208744:dw|

- UsukiDoll

so if we use difference of squares formula
|dw:1438755285695:dw|
solving for x gives x = 5, and -5

- UsukiDoll

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

- anonymous

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?

- UsukiDoll

solution set meaning which x value works...

- UsukiDoll

x = 5,-5,2

- anonymous

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

- UsukiDoll

sounds good.

- anonymous

Thanks for all the help. :3

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