One of the zeros is given for each of the following polynomials. Find the other zeros in the field of complex numbers.x^3-4x^2+6x-4.1-i is a zero

Khaleesi Herbert 2021-09-12 Answered
One of the zeros is given for each of the following polynomials. Find the other zeros in the field of complex numbers.
\(\displaystyle{x}^{{3}}-{4}{x}^{{2}}+{6}{x}-{4.1}-{i}\) is a zero

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Expert Answer

diskusje5
Answered 2021-09-13 Author has 3569 answers

We have given one roots of the polynomial \(\displaystyle{x}^{{3}}-{4}{x}^{{2}}+{6}{x}-{4}\) is 1-i then other root will be 1+i since, roots of polynomial occur in complex conjugate.
Now, we have to find just one real roots.
For check x=2 we see
\(\displaystyle{x}^{{3}}-{4}{x}^{{2}}+{6}{x}={\left({2}\right)}^{{3}}-{4}{\left({2}\right)}^{{2}}+{6}{\left({2}\right)}-{4}\)
\(\displaystyle={8}-{16}+{12}-{4}\)
\(\displaystyle={0}\)
Therefore, x=2 is a real root of the function.
Therefore, another two roots of the polynomials are 2,1+i

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