Degree-4 Polynomial Solve the equation $x^4 - 14x^3 + 50x^2 -14x + 1 = 0$.   I am not sure about how to best proceed, and would like a solution that does not involved the generalised quartic formula.
 A: Not knowing the substitution trick, you can anyway infer that if $x$ is a solution, then $1/x$ as well, so that the polynomial can be factored in two polynomials of the second degree, and these will be palindromic too:
$$x^4 - 14x^3 + 50x^2 -14x + 1 =(x^2+Ax+1)(x^2+Bx+1).$$
Developing and identifying,
$$A+B=-14,\\1+AB+1=50.$$
The solutions are $$\frac{-14\pm\sqrt{14^2-4\cdot48}}2=-8,-6.$$
Now solve
$$x^2-8x+1=0,\\x^2-6x+1=0.$$
A: A more detailed solution:
If we divide the equation by $x^2$:
$$\frac{x^4}{x^2} - \frac{14x^3}{x^2} + \frac{50x^2}{x^2} - \frac{14x}{x^2} + \frac{1}{x^2} = x^2 - 14x + 50 - \frac{14}{x} + \frac{1}{x^2}$$
Then, combining like terms, we notice that:
$$x^2 + \frac{1}{x^2} - 14\left(x+\frac{1}{x}\right) + 50$$
If we let $y = x+\frac{1}{x}$ 
Note that:
$$\left(x+\frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2$$
Therefore,
$$x^2 + \frac{1}{x^2} - 14\left(x+\frac{1}{x}\right) + 50 = y^2 -2 -14y + 50 =0$$
$$y^2 - 14y +48 =0$$
$$(y-6)(y-8) = 0$$
Therefore, $x + \frac{1}{x} = 6$ and $x + \frac{1}{x} = 8$
Can you take it from here?
A: Hint: First observe the equation is palindromic. Divide throughout with $x^2$ and rewite it as a quadratic in $\left(x+\dfrac{1}{x}\right)$.
