1
$\begingroup$

This is the first time I see this kind of problem, so it might be trivial but I am just not used to it.

What are the roots of $x^3-6ix^2-11x+6i$

I am not sure If I should ignore the imaginary numbers and simply compute the polynomial or factor the imaginary part out separately.

I tried to use the rational polynomial root test but it has no rational roots when I ignore the Imaginary coefficients.

When I factor them out as $x^3-11x-i(6x^2-6)$ I get $i$ and $-i$ as a root which is definitely wrong.

All I ask for here is to provide me advice on what method should I use to solve this type of problems. Thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ @abiessu yes I sorry, I corrected it $\endgroup$ – Fred Dec 19 '17 at 14:50
  • 1
    $\begingroup$ Hint: search for pure complex roots. $\endgroup$ – lulu Dec 19 '17 at 14:53
  • 1
    $\begingroup$ Hint: define $u$ such that $x=iu$ $\endgroup$ – Martigan Dec 19 '17 at 14:54
  • 3
    $\begingroup$ If you plug in $x=iy$, you get $-iy^3+6iy^2-11iy+6i$, which should have at least one real solution in $y$... This approach is not available in general, but is reasonable here where the even and odd degree terms all match in terms of coefficients being complex (or not). $\endgroup$ – abiessu Dec 19 '17 at 14:54
  • 2
    $\begingroup$ @abiessu I was going to put exactly the same - the alternating real and imaginary coefficients indicate this approach. $\endgroup$ – Mark Bennet Dec 19 '17 at 14:58
3
$\begingroup$

Hint:

$x=iy$

Now if $f(y)=y^3-6y^2+11y-6$

$f(1)=?$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.