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I’ve heard that a resolvent is very useful in finding the roots of the polynomial. But I’m not sure what a resolvent even is. As much as I can figure out, it’s just another polynomial. But that makes no sense, because why would you want extra work?

And how would you find the resolvent of a polynomial? For example, what’s the resolvent of a quintic?

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The resolvent of a polynomial is a polynomial of lower degree who’s roots relate to the roots of the original polynomial. Anyways, let us find the solution to the general quartic, and I will highlight the resolvent cubic.

Solve for $x$ in

$$ax^4+bx^3+cx^2+dx+e=0$$

$$x^4+b’x^3+c’x^2+d’x+e’=0\tag{divide both sides by $a$}$$

$$u^4+a_uu^2+b_uu+c_u=0\tag{Substitution $x=u-\frac{b’}4$}$$

$$u^4+nu^2+n’=mu^2+m’u+m”\tag{introduce new constants}$$

By equating parts:

$$\begin{cases}n-m=a_u\\m’=-b_u\\n’-m”=c_u\end{cases}$$

Since we want both sides to be perfect squares,

$$(u^2+\lambda_1)^2=(\sqrt mu+\lambda_2)^2$$

So that we may take the square root of both sides and take the remaining quadratic. But to do that, we must first know what $\lambda_{1,2}$ are.

To do this, expand both sides, then equate more parts.

$$u^4+2\lambda_1u^2+\lambda_1^2=mu^2+2\sqrt m\lambda_2u+\lambda_2^2$$

This gives us a whole new set of things to equate:

$$\begin{cases}2\lambda_1=n\\\lambda_1^2=n’\\2\sqrt m\lambda_2=m’\\\lambda_2^2=m”\end{cases}$$

Interestingly, if you solve by repeatedly substituting with everything we know, the solutions to $\lambda_{1,2}$ are easy, if we knew what $n$ and $m$ were.

Solving for $n,n’,m,m’,$ and $m”$ involve a cubic polynomial:

$$\boxed{(m”)^3-(4a_u^3-c_u)(m”)^2+2a_ub_u^2m”-\frac14b_u^4=0}$$

The above is a resolvent cubic that has a root equal to $m”$ with coefficients in known constants $a,b,c$.

You may derive other constants and then solve the quadratic, and follow up with lots of un-substitutions to get the final answer.

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