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January 31, 2015

January 31, 2015

Posted by **Jamie** on Monday, March 23, 2009 at 11:58am.

My answer: the possible rational roots are {±1, ±2, ±3, ±6, ±7, ±14, ±21, ±42. ±1/2, ±3/2, ± 7/2, ±21/2}.

Correct?

- Advanced Math. -
**Reiny**, Monday, March 23, 2009 at 1:08pmyes, and guess what?

x = 1 works!

after a quick synthetic division I got

2x^3 + 17x^2 + 23x - 42 = 0

(x-1)(x^2 + 19x + 42) = 0

and that quadratic factors very nicely again.

- Advanced Math. -
**Count Iblis**, Monday, March 23, 2009 at 1:18pmYes, this is what you get when you apply the Rational Roots theorem.

But your list is very long, so it is of little use. It is a bit like answering the question by saying that all possible rational roots are the members of Q (as the problem didn't specifically say to list all the roots that the Rational Roots theorem yields).

- Advanced Math. -
**Bartolo Windu**, Wednesday, October 31, 2012 at 5:30pmIncorrect

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