Let $P$ be a family of polynomials $\mathcal P:=\{p_n(x):n\in\mathbb N\}$ $$p_n(x)=x^3+2(1-n)x^2+n(n-4)x+2n^2$$ Let $f(x)\in\mathbb R[x]$ s.t. $\;\gcd(p_n(x),f(x))=d(x),\;\forall n\in\mathbb N.$
$(a)$ Are there any $f(x)\;\&\;d(x),\;\deg(d(x))=1$?
$(b)$ Are there any $f(x)\;\&\;d(x)$ s.t. $d(x)$ has a multiple root?
My attempt:
By the theorem:
If $\alpha=\frac{p}{q}\in\mathbb Q$ is a root of the equation: $$a_0x^n+a_1x^{n-1}+\ldots+a_{n-1}x+a_n=0,\;a_i\in\mathbb Z\;\forall i\in\{0,1,\ldots,n\},\;\gcd(p,q)=1$$ then $p|a_n\;\land\;q|a_0.$
Since the statements must hold $\forall n\in\mathbb N$, let $\alpha=\pm\frac{2}{1}$
Case $ii:$ For $\alpha=-2$, by long division:
$$\frac{x^3+2(1-n)x^2+n(n-4)x+2n^2}{x+2}=(x-n)^2$$
$$\implies p_n(x)=(x+2)(x-n)^2\implies x_1=-2,x_{2,3}=n$$
$(a)$ $\deg(d(x))=1\implies d(x)=x+2$
$$\implies f(x)=(x+2)q(x),\;\gcd(q(x),(x-n)^2)=1\;\;\;\;\;\;\;\;\;\;\;\;$$
$(b)$ If $d(x)$ has multiple roots $\alpha$, then must $\alpha=-2$ $$\implies f(x)=(x+2)^kq(x),\;\gcd(q(x),(x-n)^2)=1$$
Case $ii:\;\alpha=2$ isn't a root of the given equation.
My answer to both $(a)\;\&\;(b)$ is yes.
Is this correct?
Since $p_n(x)=(x-n)^2(x+2)$ the answer to (a) is yes because you can take $f(x)=x+2$.
Your answer of $d(x)=x-2$ does not work since this is not a factor of, say, $p_1(x)$.
For (b) the answer is also yes because if $f(x)=(x+2)^3$, then $d(x)=x+2$ and $f(x)=(x+2)^2 d(x)$.
[I hope this is what you are intending.]