**Problem**: (360 Problems For Mathematical Contests - 1.1.59) Solve the equation

$ x^4-(2m+1)x^3+(m-1)x^2+(2m^2+1)x+m = 0 $

where $ m $ is a real parameter.

**Solution**: Well, this equation just screams to be factored... but doing so is the challenge. Instead of a quartic in $ x $ let's write it as a quadratic in $ m $.

$ (2x)m^2-(x-1)(2x^2+x+1)m+x(x-1)^2(x+1) = 0 $

after everything is factored fully. Well this looks a little more manageable, so we guess for factors. Noticing the $ x-1 $ term on the $ m $, we can see that $ x-1 $ must appear in both factors, so the square must be split. After some work, we come up with

$ [2xm-(x+1)(x-1)][m-x(x-1)] = 0 $

$ (x^2-2mx-1)(x^2-x-m) = 0 $.

Solving this, we have either

$ x^2-2mx-1 = 0 \Rightarrow x = m \pm \sqrt{m^2+1} $

or

$ x^2-x-m = 0 \Rightarrow x = \frac{1 \pm \sqrt{1+4m}}{2} $,

both just from using the quadratic formula. Hence these are our solutions for $ x $ in terms of the parameter $ m $. QED.

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Comment: This technique of switching the variable is very useful because solving quadratics is much simpler than higher degree polynomials. Problems will often give the equation in a form like this one with one variable of a large degree and another of degree two. It's possible to discover the factorization by looking at the fourth degree polynomial, but much more difficult.

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Practice Problem: (360 Problems For Mathematical Contests - 1.1.60) Solve the equation

$ x^{2n}+a_1x^{2n-1}+\cdots+a_{2n-2}x^2-2nx+1 = 0 $

if all of its roots are positive reals. [Note: Original problem omitted positive, but I think it's necessary from looking at the solution.]

Practice Problem: ARML 2005 Individual #8 :D

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