How to find the rational root in this case. Is there a relationship between all the roots?

Question

I know that the complex roots come in pairs and I know the Vieta's formulas. But I still don't know how to deal with this problem.

Answer 1

Since $2+i$ is a root of $g(z),g'(z),g''(z),$ it is a triple root of $g(z).$ As you pointed out, due to real coefficients is also $2-i$ a root of each of $g(z),g'(z),g''(z),$ therefore it is a triple root. Compute

$$(z-(2+i))(z-(2-i))=z^2-4z+5.$$

Thus $$g(z)=(z^2-4z+5)^3(z-r),$$ where $r$ is the real root. The free coefficient is a product of the free coefficients from the above factored form. Hence you have $5^3(-r)=-53$ and $$r=\frac{53}{125}$$ is the real root.

Answer 2

Hint:

(a) $2\pm i$ is a triple root of $g$ and so $g(z)=(z-2+i)^3(z-2-i)^3h(z)$, with $h$ of degree $1$.

(b) $-a_6$ is the sum of the roots, counted with multiplicity.