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December 20, 2014

December 20, 2014

Posted by **Karlee** on Tuesday, March 26, 2013 at 6:08pm.

f(x)=2x^3-x^2+2x-1

- Use the rationals theorem -
**Reiny**, Tuesday, March 26, 2013 at 6:10pmfollow the same steps I just showed you in your last post

http://www.jiskha.com/display.cgi?id=1364332046

- Use the rationals theorem -
**Count Iblis**, Tuesday, March 26, 2013 at 8:28pmI 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.

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