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1 - 6i is a zero of f(x) = x^4 - 2x^3 + 38x^2 - 2x + 37. How would I find another zero?
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Imaginary zeros always come in pairs, I believe, in the form \[a \pm bi\] So you can get another zero by just slightly modifying the one you know!
it is the conjugate what @BangkokGarrett said
I was thinking that but it seemed to easy and I have to explain the process in how I found it, I'm not sure if they would accept me using 1+6i
or am I just overthinking it
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