anonymous
  • anonymous
Can someone help me out with Rational Zero Theorem problems? x^3 - 2x^2 - 25 + 50 = 0 https://i.imgur.com/NLjwYRX.png
Mathematics
schrodinger
  • schrodinger
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UsukiDoll
  • 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
  • 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
  • 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
  • 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
  • UsukiDoll
12 numbers to test out .......
anonymous
  • 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
  • anonymous
I'm getting -5, 5, and 2, but those are probably waaay off.
UsukiDoll
  • 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
  • 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
  • UsukiDoll
either way ... it's right.
UsukiDoll
  • UsukiDoll
but I would've done long division... even though it takes forever.
freckles
  • freckles
you can check @mintalicarth by finding the following product (x-5)(x+5)(x-2)
UsukiDoll
  • 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
  • UsukiDoll
hmmm it's been a while since I've done synthetic... maybe I should do that :P |dw:1438755208744:dw|
UsukiDoll
  • UsukiDoll
so if we use difference of squares formula |dw:1438755285695:dw| solving for x gives x = 5, and -5
UsukiDoll
  • 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
  • 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
  • UsukiDoll
solution set meaning which x value works...
UsukiDoll
  • UsukiDoll
x = 5,-5,2
anonymous
  • 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
  • UsukiDoll
sounds good.
anonymous
  • anonymous
Thanks for all the help. :3

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