smc

Let the other quadratic be $x^2+bx+c$, where $(x^2+2x+5)(x^2+bx+c)=x^4+p{x}^2+q$. Multiply the two quadratics to get $$x^4+(b+2)x^3+(c+2b+5)x^2+(2c+5b)x+5c$$ Since $x^4+p{x}^2+q$ have no $x^3$ term and no $x$ term, the coefficients of these terms must be zero. $$b+2=0$$ $$b=-2$$ $$2c+5b=0$$ $$2c-10=0$$ $$c=5$$ Substitute $c$ and $b$ back to get $$x^4+6x^2+25$$ Thus, $p=6$ and $q=25$, so the answer is $\boxed{6,25}$.