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The limiting polynomial has a double root at $x=-1$, so two roots require a Puiseux series. For positive $ϵ$, put $\delta =\sqrt{ϵ}$ and $x=-1+a\delta +b{\delta }^2+\cdots$. Substitution and dominant balance giveThus $a=\pm 1$ and $b=-3/2$. The two finite roots have the three-term asymptotic expansionsThis is square-root splitting of a double polynomial root. For negative $ϵ$, the same formulas apply with a chosen complex square root and describe a complex conjugate pair.

The third root is a divergent algebraic root in a singular perturbation. Set $x=X/ϵ$: the equation for $X$ is $X^3+X^2+2ϵX+{ϵ}^2=0$. Its nonzero limiting root is $X=-1$. Substituting $X=-1+Aϵ+B{ϵ}^2+\cdots$ yields $A=2$ and $B=3$, henceAs an independent check, Vieta formulas give the sum of roots as $-1/ϵ$: the two finite-root expansions sum to $-2-3ϵ+O({ϵ}^2)$, giving the same large-root expansion.

Suppose a perturbed polynomial has the local form $A(x-x_0{)}^2+ϵB+O((x-x_0{)}^3,ϵ(x-x_0),{ϵ}^2)$ with $AB\ne 0$. Dominant balance gives $x-x_0={ϵ}^{1/2}z$ and $A{z}^2+B=0$. The two resulting Puiseux series split a double root; their leading displacements can be real or a complex conjugate pair. Ordinary Taylor series in $ϵ$ cannot describe this splitting.