Posted by Karlee on Tuesday, March 26, 2013 at 6:08pm.
follow the same steps I just showed you in your last post
http://www.jiskha.com/display.cgi?id=1364332046
I like this variant which makes use of candidates that turn out not to be a zero. This works as follows.
The possible zeroes are:
x = -1,1,1/2,-1/2 (1)
We have:
f(1) = -2
If we put
g(t) = f(1+t)
then the coefficient of t^3 is 2 and the constant term is f(1) = -2, the possible zeroes are thus:
t = +/-1 , +/-2, +/- 1/2
The possible zeroes of f are thus:
x = 1+t = 0,2,-1,3,1/2,3/2
But since all the possible zeroes are listed in (1), we can strike out the elements that are not on that list. We are thus left with:
x = -1,1/2
Then since -1 is not a zero, the only possible rational zero is 1/2, which is indeed a zero. Then you can proceed by dividing f(x) by x-1/2 and find the zeroes of the quadratic.