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Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
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Look for a simple substitution that eliminates the quadratic term of 4 x^3+5 x^2-x+2.
Eliminate the quadratic term by substituting y = x+5/12:
29/12+5 (y-5/12)^2+4 (y-5/12)^3-y = 0
That is definitely not one of the options in this multiple choice question.
Let me give the possible answers: (all negative and positive of this number)
a. 1, 1/2, 1/4, 2
b. 1, 1/2, 2, 4
c. 1, 1/2, 1/4
d. 1, 1/4, 2
u have an easy way of solving....firstly...note that a cubic eqn has atmost two complex sol..and atleast 1 real soln....so go for the real soln by trial and error...then u hav a factor...divide the polynomial by the factor...u have a quad eqn now as the quotnt....solve it to get the complex roots
@pathosdebater I'm not sure, I got similar answers (which is why I came here). There may be a typo in the question (not what I typed; the actual question itself), but no one I know has said anything about it. :\
do u mind if i say that the options, all of them are wrong..? its so coz...u cant have three complex roots for a REAL cubic eqn...complex roots come in pairs...so...
@amriju I don't mind, I found it to be that way too when I substituted it in. I don't know. Not my practice, it was the teachers. I got all of the other questions on finding zeros, this one was just confusing.
u really cant have an option with more than two values...that settles it...as for the solution u got to try this way...if a,b,c be the roots then\[\sum_{?}^{?} a=-5/4...\sum_{?}^{?} ab=1/4....abc=-2/4=-1/2\]