We can factorize the quadratic as
[tex]x^2 + 3x + 2 = (x - \alpha) (x - \beta)[/tex]
and expanding the right side leads to
[tex]x^2 + 3x + 2 = x^2 - (\alpha + \beta) x + \alpha \beta \implies \begin{cases} \alpha + \beta = -3 \\ \alpha \beta = 2 \end{cases}[/tex]
The polynomial we want will have the factorization and expanded form
[tex]\left(x - \left(1 + \dfrac\beta\alpha\right)\right) \left(x - \left(1 + \dfrac\alpha\beta\right)\right) = x^2 - \left(2 + \dfrac\beta\alpha + \dfrac\alpha\beta\right) x + \left(1 + \dfrac\beta\alpha\right) \left(1 + \dfrac\alpha\beta\right)[/tex]
and notice that the constant term is actually the same as (but has the opposite sign of) the coefficient of the [tex]x[/tex] term :
[tex]\left(1 + \dfrac\beta\alpha\right) \left(1 + \dfrac\alpha\beta\right) = 1 + \dfrac\beta\alpha + \dfrac\alpha\beta + 1 = 2 + \dfrac\beta\alpha + \dfrac\alpha\beta[/tex]
Rewrite the coefficient as
[tex]2 + \dfrac\beta\alpha + \dfrac\alpha\beta = 2 + \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = 2 + \dfrac{\alpha^2 + \beta^2}2[/tex]
Now, observe that
[tex](\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \implies \alpha^2 + \beta^2 = (-3)^2 - 2^2 = 5[/tex]
Then the coefficient is simply [tex]2 + \frac52 = \frac92[/tex], so the polynomial we want is
[tex]\boxed{x^2 - \dfrac92 x + \dfrac92}[/tex]
