I have a problem with a multiple choice question. I feel that the correct answer is not among the options ad there is also not a option that states one of the above...
The question is: Which of the following sets contains only possible rational zeros of
P(x) = 2x^4 - x^3 + 6x^2 - 5x + 8

A) +- 1/4, +- 1/2, +- 1, +- 2
B) +- 1/8, +-1/4, +- 1/2, +- 1, +- 2
C) +- 1/8, +- 1, +- 2, +-4, +-8
D) +- 1/2, +- 2, +- 8
E) +-1, +- 2, +-4, +- 8, +- 16

This options confuse me as the answer I get don't correspond with any of the above options.
Am I mistaken?

Question

I have a problem with a multiple choice question. I feel that the correct answer is not among the options ad there is also not a option that states one of the above...
The question is: Which of the following sets contains only possible rational zeros of
P(x) = 2x^4 - x^3 + 6x^2 - 5x + 8

A) +- 1/4, +- 1/2, +- 1, +- 2
B) +- 1/8, +-1/4, +- 1/2, +- 1, +- 2
C) +- 1/8, +- 1, +- 2, +-4, +-8
D) +- 1/2, +- 2, +- 8
E) +-1, +- 2, +-4, +- 8, +- 16

This options confuse me as the answer I get don't correspond with any of the above options.
Am I mistaken?

hartnn · Answer

i did not get any rational or real 0 , all complex

hartnn · Answer

so maybe u are right and there is mistake in question.

lgbasallote · Answer

the answer IS within the choices...

hartnn · Answer

so u got real 0's lg ?

lgbasallote · Answer

oh yes. i did...

hartnn · Answer

wolf don't agree with u :(

lgbasallote · Answer

why? did i say anything yet?

hartnn · Answer

u meant u got real 0's , right?

lgbasallote · Answer

yes..i get real 0's

hartnn · Answer

or u got 'possible' real o's ?

lgbasallote · Answer

i got that too

hartnn · Answer

ok lg, so here it is: there are actually no real solutions to this.... but rational root theorem gives "possible" rational roots to an equation, so yes the answer IS in the choices as u said, but there are no real solutions....

lgbasallote · Answer

just messing with you math lover. no need to go "meticulous" like mathematicians

hartnn · Answer

$\frac{p}{q}=\frac{8}{2}=\pm \frac{1,2,4,8}{1,2}$
so possible roots are that given in option B
ask if u still have doubts @CHEMMIB

lgbasallote · Answer

so why did you say the answer is not in the options @hartnn ?

lgbasallote · Answer

and it should be D...not B

hartnn · Answer

oh, the first few comments ?? at that time i couldn't think of that theorem....then i realised that rational root theorem can be used here because of the word 'possible'

lgbasallote · Answer

then you went on an argument with me?

hartnn · Answer

oh, yes its D , so sorry.... and sorry for the argument.... i thought u really got real 0's and i was curious how....

lgbasallote · Answer

well the real 0 part i was messing with you...i didn't really solve for that

lgbasallote · Answer

i just solved the possible roots

anonymous · Answer

Hi guys thank you for your comments on this question. I do appreciate it a lot.
I am still uncertain about this however. If this wasn't a multiple choice question, I would've said that all the POSSIBLE zeros are: +-1, +- 2, +- 4, +- 8 and +-1/2.
Because that's the combination I get by taking all the factors of the constant term (8) over the 2 factors of the leading term (2).
Is this incorrect?
I see that you say option D is the correct choice, which is +-1/2, +-2 and +-8. I do agree with this possibilities but what about +-1 and +- 4?

Thank you!

hartnn · Answer

since that was the equation of degree 4 only, so they chose only 4 possibilities from all to give in options....

anonymous · Answer

I understand, even though option D only states 3 possibilities?
+- 1/2 ( as in +- 0.5),  +- 2,   and +- 8

hartnn · Answer

hmm....but that made it 6 0's ! :P

hartnn · Answer

anyways, u got the concept, thats enough....