Find the sum of all the integer solutions of  $x^4 - 25x^2 + 144 = 0$.
I notice that this is a quartic equation, but it has only even powers of x. That makes me think of substituting $u = x^2$, so that the equation becomes $u^2 - 25u + 144 = 0$. This is a quadratic equation, which I can solve by factoring. I look for two numbers that multiply to 144 and add to -25. I see that -9 and -16 work, so I can write $u^2 - 25u + 144 = (u - 9)(u - 16)$. Therefore, $u = 9$ or $u = 16$. But $u = x^2$, so I need to take the square root of both sides. I get $x = \pm 3$ or $x = \pm 4$. These are the four integer solutions of the original equation. To find their sum, I just add them up: $3 + (-3) + 4 + (-4) = 0$. 