can anyone show me how to do? degree 3; zeros 3;-i;8+i?
is it really zeros of \(\{3,-i,8+i\}\)?
because if so your polynomial will have complex coefficients, not real ones if it had real ones they have to come in conjugate pairs, like \(\{i,-i\}\) and \(\{8+i,8-i\}\)
I need to find all remaining zeros
if you do you will have 5 zeros, so it will not be a polynomial of degree 3, it will be a polynomial of degree 5
complex zeros of real polynomials come in conjugate pairs, if \(a+bi\) is a zero, then so is \(a-bi\)
in your case if \(-i\) is a zero then so is \(i\) and if \(8+i\) is a zero then so is \(8-i\)
but as i said, the polynomial will be degree 5 not 3
ok what if I would have degree 3; zeros 8, 8-i?
Actually the equation has complex cofficients .
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