Convert fourth degree polynomial to quadratic

Question

We have this equation :

$ax^4 + bx^3 + cx^2 + bx + a = 0$

How we can convert this equation to $$f(t)=0$$ where $f(t)$ is a polynomial of degree $2$ in $t$ ? And what's the relation between $x$ and $t$ ?

Answer 1

If $a=0$ then the equation reduces to $x \cdot (b x^2 + c x + b) = 0$ which is already factored into linear plus quadratic terms.

Otherwise $x = 0$ is not a solution, and dividing the equation by $x^2 \ne 0$ results in:

$$a(x^2+\frac{1}{x^2}) + b(x+\frac{1}{x})+c = 0$$

Then the substitution $t=x+\frac{1}{x}$ (which implies $t^2=x^2 + 2 + \frac{1}{x^2}$ thus $x^2 + \frac{1}{x^2} = t^2-2$) gives the quadratic in $t$:

$$at^2 + bt+c-2a = 0$$