Average of the zeroes

Given that

$\begin{aligned} f(x) & = x^3-3x^2 + x + 2 \\ (x-2)g(x) - 2x + 4 & = x^3-3x^2 + x + 2 \\ (x-2) g(x) & = x^3 - 3x^2 + 3x - 2 \\ \implies g(x) & = \frac {x^3-3x^2+3x -2}{x-2} \\ & = x^2 - x + 1 \end{aligned}$

By Vieta's formula , the sum of two zeroes of $g(x)$ is $1$ (negative of the coefficient of $x$ }. Therefore, the average of the two zeroes is $\frac 12 = \boxed{0.5}$ .

$\begin{aligned} f(x)&=x^3-3x^2 +x+2 \\ f(x) &= g(x) \cdot q(x) + r(x) \\ x^3-3x^2 +x+2 &= g(x) \cdot (x+2) + (4-2x) \\ x^3-3x^2 +3x-2 &= g(x) \cdot (x+2) \\ (x-2)(x^2 - x -1) &= g(x) \cdot (x+2) \\ (x^2 - x -1) &= g(x) \\ \\ \text{Sum of roots of quadratic expression } (g(x)) &= -(-1) = 1 \\ \\ \text{Average} &= \frac{\text{Sum of both roots}}{2} \\ \text{Average} &= \boxed{\frac{1}{2}} \end{aligned}$