Find the Complex Roots: 4x3 + 5x2 – x + 2. Thank you!

- anonymous

Find the Complex Roots: 4x3 + 5x2 – x + 2. Thank you!

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

try using the quadratic formula

- anonymous

alright. let me try this out.

- anonymous

ok take your time

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

how would that help?
i don't understand.

- anonymous

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

- anonymous

@hartnn ?

- anonymous

do you have an easier way of explaining this?

- hartnn

let me think...

- anonymous

##### 1 Attachment

- anonymous

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

- campbell_st

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

One second

- anonymous

Sorry for the confusion.

- anonymous

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

##### 1 Attachment

- 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

- anonymous

@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. :\

- anonymous

What textbook is it?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I got it! It is definitely C

- 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...

- anonymous

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

- anonymous

2 is not a complex root

- anonymous

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

- amriju

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

- anonymous

@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.

- amriju

have u done complex numbers @pathosdebater

- anonymous

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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.