Find the remaining zeros of the polynomial:

\[ x^3 - 6x^2 + 36x - 218 \]

Given zero: \(-6i\)

x 3 − 6 x 2 + 36 x − 216 = x 2 ( x − 6 ) + 36 ( x − 6 ) = ( x − 6 ) ( x 2 + 36 ) = = ( x − 6 ) [ x 2 − 36 ⋅ i 2 ) = ( x − 6 ) ( x − 6 i ) ( x + 6 i ) ( x − 6 ) ( x − 6 i ) ( x + 6 i ) = 0 ⇔ x − 6 = 0 or x − 6 i = 0 or x + 6 i = 0 . x = 6 x = 6 i x = − 6 i

Answered by kate200468 | 2024-06-10

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 '.

Answered by AL2006 | 2024-06-10

The remaining zeros of the polynomial x 3 − 6 x 2 + 36 x − 218 , given one zero as − 6 i , are 6 i and 6 . The polynomial factors to reveal these zeros effectively. Thus, the complete set of zeros is − 6 i , 6 i , and 6 .
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Answered by kate200468 | 2024-10-02