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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.

A regular perturbation admits an asymptotic expansion about the limiting problem on the relevant scale without losing its solutions or conditions. A singular perturbation changes the limiting structure, so some solutions require a different scale: a differential equation may lose an order, or a polynomial may lose its degree.

The two finite roots approach $s=\pm \sqrt{2}$. Set $x=s+ϵa+O({ϵ}^2)$. Substitution into the polynomial gives $s^2-2=0$ at leading order and $s^3+2sa=0$ at the next order. Since $s^2=2$, $a=-1$ for either sign. Each branch is a regular perturbation of a simple algebraic root, withThe limiting quadratic has only two roots, whereas the original equation has three. The missing root is a divergent algebraic root in a singular perturbation. The dominant balance for algebraic roots sets $x={ϵ}^{-1}X$, giving $X^3+X^2=2{ϵ}^2$. Its nonzero leading root is $X=-1$. Put $X=-1+a{ϵ}^2+O({ϵ}^4)$; the left side is $a{ϵ}^2+O({ϵ}^4)$, so $a=2$. Hence the second nonzero term occurs at order $ϵ$, not order one:For small nonzero real $ϵ$ these three branches are real and distinct; the divergent branch changes sign with $ϵ$. The two bounded branches are regular even though the full root problem is singular.