An algebra problem by Sakanksha Deo

The solution can be found using long division as follows:

We note that for $x^2+2x+5$ to be a factor of $x^4+px^2+q$ ,

$\begin{cases} 2(p-1)=10 & \Rightarrow p-1 = 5 &\Rightarrow p = 6 \\ q = 5(p-1) & \Rightarrow q = 5\times 5 &\Rightarrow q = 25 \end{cases} \quad \Rightarrow p, q = \boxed {6,25}$

let the roots of $x^2+2x+5$ be $r_1,r_2$ . than the following must also be roots of $x^4+px^2+q$ .it means $\begin{aligned} r_1^4+pr_1^2+q=0\\ r_2^4+pr_2^2+q=0 \end{aligned}$ adding these 2, $(r_1^4+r_2^4)+p(r_1^2+r_2^2)+2q=0$ by vieta's we have $r_1+r_2=-2,r_1r_2=5$ hence $r_1^4+r_2^4=((r_1+r_2)^2-2r_1r_2)^2-2r_1^2r_2^2=-14$ $r_1^2+r_2^2=(r_1+r_2)^2-2r_1r_2=-6$ hence, $-14-6p+2q=0\rightarrow p=3q+7$ the only ones from option satisfying this is $\boxed{6,25}$