baonhulovely
Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

find the remaining zero x^3-6x^2+36x-218 zero:-6i

Sagot :

kate200468
[tex] x^3-6x^2+36x-216=x^2(x-6)+36(x-6)=(x-6)(x^2+36)=\\\\=(x-6)[x^2-36\cdot i^2)=(x-6)(x-6i)(x+6i)\\\\(x-6)(x-6i)(x+6i)=0\\\\\Leftrightarrow\ \ \ x-6=0\ \ \ \ or\ \ \ \ x-6i=0\ \ \ \ or\ \ \ \ x+6i=0\\\\.\ \ \ \ \ \ x=6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=6i\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-6i[/tex]
AL2006
I know this isn't fair, but I happen to know that imaginary or complex roots always
occur in conjugate pairs.  So if -6i is a root, then +6i also must be one.

The expression also has one real root.

It's between 6.02765006 and 6.02765007 .

The reason for that is probably a mis-type or mis-copy in the question.
The ' 218 ' at the end was probably supposed to be ' 216 '.  In that case,
the real root would have been exactly ' 6 '.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.