Will factor theorem help here?

Using the more familiar $x$ and $f(x)$ , we have:

$\begin{aligned} f(x) & = \eta x^4+\delta x^3 + 32x^2-17x + 6 \end{aligned}$

Since $(3x^2-2x+1)|f(x)$ , we can assume:

$\begin{aligned} f(x) & = (3x^2-2x+1)(ax^2+bx+c) \end{aligned}$

Comparing the coefficients, we have:

$\begin{cases} 3a & = \eta & ... (1) \\ 3b-2a & = \delta & ... (2) \\ a - 2b +3c & = 32 & ... (3) \\ b -2c & = -17 & ... (4) \\ c & = 6 & ... (5) \end{cases}$

From Eq. 3: $c=6$ , substituting it in Eq. 4: $\Rightarrow b = -5$

Substituting $b= -5$ and $c = 6$ in Eq. 3: $\Rightarrow a = 4$

From Eq. 1: $\eta = 3a = 12$

From Eq. 2: $\delta = 3b-2a = -23$

Therefore, $\eta + \delta = 12 - 23 = \boxed{-11}$

We use the trick of forcing the factor.

$\eta\gamma^4+\delta\gamma^3+32\gamma^2-17\gamma + 6 \\ \Rightarrow \eta\gamma^4+\delta\gamma^3 +14\gamma^2+18\gamma^2-12\gamma-5\gamma+6 \\ \Rightarrow\eta\gamma^4+\delta\gamma^3+14\gamma^2-5\gamma+18\gamma^2-12\gamma+6 \\ \Rightarrow \eta\gamma^4+\delta\gamma^3+14\gamma^2-5\gamma+6(3\gamma^2-2\gamma+1) \\\Rightarrow \eta\gamma^4+(\delta+15)\gamma^3+4\gamma^2-15\gamma^3+10\gamma^2-5\gamma+6(3\gamma^2-2\gamma+1) \\ \Rightarrow\eta\gamma^4+(\delta+15)\gamma^3+4\gamma^2-5\gamma(3\gamma^2-2\gamma+1)+6(3\gamma^2-2\gamma+1) \\ \Rightarrow 4\gamma^2\left(\dfrac{\eta}{4}\gamma^2+\dfrac{(\delta+15)}{4}\gamma+1\right)-5\gamma(3\gamma^2-2\gamma+1)+6(3\gamma^2-2\gamma+1)$

For $(3\gamma^2-2\gamma+1) \ | \ \zeta(\gamma)$ ,

$\dfrac{\eta}{4}\gamma^2+\dfrac{\delta+15}{4}\gamma+1 = (3\gamma^2-2\gamma+1) \\ \Rightarrow\dfrac{\eta}{4}=3 \ , \ \dfrac{\delta+15}{4}=-2 \\ \Rightarrow \eta=12 \ , \ \delta=-23 \\ \Rightarrow \eta+\delta=12-23=\boxed{-11}$

Indeed the other factor is $4\gamma^2-5\gamma+6$ .