jmc

We have that $$(x^2+2x+5)(x^2+bx+c)=x^4+P{x}^2+Q.$$for some coefficients $b$ and $c.$ Expanding, we get $$x^4+(b+2)x^3+(2b+c+5)x^2+(5b+2c)x+5c=x^4+P{x}^2+Q.$$Matching coefficients, we get $$\begin{array}{rl}b+2& =0,\\ 2b+c+5& =P,\\ 5b+2c& =0,\\ 5c& =Q.\end{array}$$Solving $b+2=0$ and $5b+2c=0,$ we get $b=-2$ and $c=5.$ Then $P=2b+c+5=6$ and $Q=5c=25,$ so $P+Q=\boxed{31}.$