Originally Posted by

**earboth** Hello topsquark,

Thanks for your reply!

first: It wasn't me who "invented" this method.

second: The following method only works if:

- the coefficients are arranged(?) symmetrically in the equation **and**

- $\displaystyle x \neq 0$

Then you can divide the equation by x²:

$\displaystyle x^4+2x^3-13x+2x+1=0$ Divide by x²

$\displaystyle x^2+2x-13+\frac{2}{x}+\frac{1}{x^2}=0$ Re-arrange:

$\displaystyle x^2+\frac{1}{x^2}+2x+\frac{2}{x}-13=0$ Factor:

$\displaystyle x^2+\frac{1}{x^2}+2 \left( x+\frac{1}{x} \right)-13=0$

Now set $\displaystyle z = x+\frac{1}{x}$. Then

$\displaystyle z^2 = x^2+2+ \frac{1}{x^2}$. That means

$\displaystyle z^2 -2= x^2+ \frac{1}{x^2}$

And now you can use substitution. The equation becomes:

$\displaystyle z^2-2+2 \cdot z-13=0$ which will give z = -5 or z = 3.

Re-substituate to solve for x

The actual solution for x doesn't matter. I only wanted to show a very special method to solve those equations.

EB