Polynomi with complex coeffiecients

[Liene]

Hi, again!
I have another problem!
I have polynomial :
2ix^3+(2+i)x^2+(1+i)x+1

and I know that one root is i.
I have to find 2 others.

I came to idea to factor this polynomial with x-i;
But I get some unbelievable result like 2xi+(2+i)x+(-2+3i)R-2i-2
I understand that I just dont understand the method of long division by complex numbers.
Long division calculators shows me quite short complex roots - not like this.
Can anyone try to solve this.
I just want learn the method to solve it.

 

[daon2]

What is R?? When dividing by \(\displaystyle x-i\), the result should be \(\displaystyle 2ix^2+ix+i = (2x^2+x+1)i\). I will start it for you.

\(\displaystyle \hspace{1cm} \color{red}{+ 2ix^2} \color{blue}{+ ix} \\
x-i\,\,\overline{)\,2ix^3+(2+i)x^2+(1+i)x+1}\\
\hspace{0.8cm}\underline{\color{red}{-(2ix\,\,\,\,\,\,\,\,\,\,\,\,\,+ 2x^2)}}\\
\hspace{3.5cm}ix^2\\
\)

Then the quadratic formula will serve you the other two.

 

[Novice]

If one root is i then the other is -i as complex roots occur in pairs. Hope this helps.

 

[HallsofIvy]

If one root is i then the other is -i as complex roots occur in pairs. Hope this helps.

No, it doesn't help because it is wrong! If a polynomial equation with real coefficients has a specific complex number (such as i) as a root, then it also has its complex conjugate (-i) as a root.

That does not apply here because the coefficients are not real numbers.

 

[Novice]

No, it doesn't help because it is wrong! If a polynomial equation with real coefficients has a specific complex number (such as i) as a root, then it also has its complex conjugate (-i) as a root.

That does not apply here because the coefficients are not real numbers.

Oops! Sorry!