# How do the roots of a polynomial change when another polynomial is added?

I need to obtain an analytical solution to an equation of the following form: $$(x-a)(x-b)(x-c)=d(x-e)(x-f),$$ where $a$, $b$, $c$, $d$, $e$, and $f$ are known numbers and $x$ is the variable.

Of course, the equation can be reduced to a "simple" equation for the roots of a 3rd-order polynomial, and its solution is provided by the Cardano formula (and other similar methods), but the result is too complex...

So the question arises: does any simpler method exist that makes use of the known roots of the "constituent polynomials" (left-hand and right-hand side of my equation)?

P.S. It is possible to try the Cardano method with the polynomial coefficients expressed through the roots $a$...$c$,$e$,$f$ and then to simplify the resulting gargantuan formulae using a CAS (computer algebra system) hoping that the CAS will cope with it... But I am not so sure that the CAS is powerful enough in the dark art of formulae simplification.

• Suppose you want to find the roots of $x^3-x^2+4=0.$ It does not get get any easier (or harder) if you rewrite as $(x-0)(x-0)(x-0)=(x+2)(x-2).$ – Aaron Meyerowitz Apr 17 '16 at 22:27

As you point out, this is no harder than solving a general cubic. It turns out that it is also no easier.

For any choice of the $6$ parameters $a,b,c,d,e,f$ your desired equation can be multiplied out to a cubic $x^3+Ax^2+Bx+C$ which can be solved by Cardano's formula. You hope for a nicer solution perhaps based on the form. However one might expect that with $6$ parameters one can easily get $A,B,C$ to be anything desired. This turns out to be the case.

Here is one way assuming all parameters constrained to be real. The complex case is as easy.

Pick $a=b=0$ and $c=-(A+d)$ where $d$ is $+1$ or $-1$ according as $C \ge 0$ or $C \lt 0.$ Then $e$ and $f$ are the (real) roots $$\frac{Bd \pm \sqrt{B^2+4|C|}}{2}$$ of the quadratic equation. $$x^2-\frac{B}{d}x-\frac{C}{d}.$$

While a simple closed-form solution isn't available, you could try series expansions in the parameter $d$, if $d$ is small. Thus suppose you want the root that is $a$ at $d=0$. For simplicity, rewrite the equation as $$X (X-B)(X-C) = d(X-E)(X-F)$$ where $X=x-a$, $B = b-a$, $C=c-a$, $E=e-a$, $F=f-a$. Then this root is \eqalign{ X &= {\frac {EF}{BC}}d+ \left( -{\frac {{E}^{2}F}{{B}^{2}{C}^{2}}}-{\frac {E{F}^{2}}{{B}^{2}{C}^{2}}}+{\frac {{E}^{2}{F}^{2}}{{B}^{2}{C}^{3}}}+{ \frac {{E}^{2}{F}^{2}}{{B}^{3}{C}^{2}}} \right) {d}^{2}\cr+ &\left( { \frac {{E}^{3}F}{{B}^{3}{C}^{3}}}+3\,{\frac {{E}^{2}{F}^{2}}{{B}^{3}{C }^{3}}}+{\frac {E{F}^{3}}{{B}^{3}{C}^{3}}}-3\,{\frac {{E}^{3}{F}^{2}}{ {B}^{3}{C}^{4}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{3}{C}^{4}}}+2\,{ \frac {{E}^{3}{F}^{3}}{{B}^{3}{C}^{5}}}-3\,{\frac {{E}^{3}{F}^{2}}{{B} ^{4}{C}^{3}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{4}{C}^{3}}}+3\,{\frac {{ E}^{3}{F}^{3}}{{B}^{4}{C}^{4}}}+2\,{\frac {{E}^{3}{F}^{3}}{{B}^{5}{C}^ {3}}} \right) {d}^{3}\cr \cr+&\ldots} This (and as many terms as you want) can be obtained from the Lagrange–Bürmann formula.

The short answer is no. -- Take e.g. the case $a = b = c = 0$ and $d = e = f = 1$. Then the polynomial whose roots you need to find is $x^3-x^2+2x-1$, which has Galois group ${\rm S}_3$.

The same works for higher degrees as well, e.g. in order to solve the equation $$(x-0)(x-0)(x-0)(x-0)(x-0) \ = \ (x-1)(x-1),$$ you need to find roots of the polynomial $x^5-x^2+2x-1$ whose Galois group is not solvable and whose roots hence cannot be written as radicals.

Perhaps in some cases, you can apply Rouché's Theorem