## radh 3 years ago Find the Complex Roots: 4x3 + 5x2 – x + 2. Thank you!

1. NickR

alright. let me try this out.

3. NickR

how would it work considering the first term is cubed, not squared?

5. NickR

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

how would that help? i don't understand.

i know it simplifies to 4y3 - 37y/12 + 2.99.

8. NickR

@hartnn ?

9. NickR

do you have an easier way of explaining this?

10. hartnn

let me think...

11. pathosdebater

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

13. campbell_st

lol.. wow thats neat work... @pathosdebater

14. pathosdebater

Is the problem 4x3 + 5x2 – x + 2 or 4*3 + 5*2 – x + 2? (Just making sure)

It's exponential, yeah. 4x^3 + 5x^2 - x + 2.

16. pathosdebater

What? Oh... I didn't know that. I thought it was 4x*3 + 5x*2 - x+2

17. pathosdebater

One second

Sorry for the confusion.

19. pathosdebater

This is so strange. I keep getting irrational answers. Is there anything wrong in my calculations which could be causing this?

20. amriju

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. :\

22. pathosdebater

What textbook is it?

It's not a textbook, it's a practice given by the teacher. (I searched online for duplications but it looks original.)

24. pathosdebater

Hmm... well there are definitely only 3 roots, so that narrows it down to C and D.

Thanks for the help though, I'm just going to guess. I appreciate the effort.

26. pathosdebater

I got it! It is definitely C

27. amriju

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...

That was my initial guess, but can you explain how?

29. pathosdebater

2 is not a complex root

30. pathosdebater

Let me see if I can try and explain...

31. amriju

its got to have two complex roots ..if not zero

@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.

33. amriju

have u done complex numbers @pathosdebater

34. pathosdebater

Only briefly. I'm not a wizard at them... yet

35. amriju

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$\sum_{?}^{?} a=-5/4...\sum_{?}^{?} ab=1/4....abc=-2/4=-1/2$