### Instruction: If $x^2 + 2x + 5$ is a factor of $x^4 + Px^2 + Q,$ find $P + Q.$.### Response: I notice that the factor has degree 2, so the other factor must also have degree 2.
I can write the other factor as $x^2 + ax + b$ for some constants $a$ and $b.$
Then I have $(x^2 + 2x + 5)(x^2 + ax + b) = x^4 + Px^2 + Q.$
I can expand the left-hand side using the distributive property and collect like terms.
I get $x^4 + (a + 2)x^3 + (b + 2a + 5)x^2 + (5a + 2b)x + 5b.$
Now I can compare the coefficients of each power of $x$ on both sides.
The coefficient of $x^4$ is 1 on both sides, so that's consistent.
The coefficient of $x^3$ is 0 on the right-hand side, so I must have $a + 2 = 0.$
That means $a = -2.$
The coefficient of $x^2$ is $P$ on the right-hand side, so I must have $P = b + 2a + 5.$
Substituting $a = -2,$ I get $P = b - 4 + 5 = b + 1.$
The coefficient of $x$ is 0 on the right-hand side, so I must have $5a + 2b = 0.$
Substituting $a = -2,$ I get $-10 + 2b = 0.$
That means $b = 5.$
The constant term is $Q$ on the right-hand side, so I must have $Q = 5b.$
Substituting $b = 5,$ I get $Q = 25.$
Now I have $P = b + 1 = 5 + 1 = 6$ and $Q = 25.$
The question asks for $P + Q,$ so I add them up.