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The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

Not near the minimum at $1$. The derivative being set to zero factors asso $1$ has multiplicity of a root equal to three. The ordinary Newton iteration at a multiple root is generally only linearly convergent, with error ratio $(3-1)/3=2/3$ here.

In fact, cancellation in $x-f^{\prime }(x)/f^{\prime \prime }(x)$ gives the exact error identity, away from the original undefined quotient at the root,For iterates approaching $1$, the factor tends to $2/3$, proving linear convergence. The other zero of $f^{\prime }$ at $-1$ is simple and could instead give quadratic local convergence; the failure at $1$ is due to its multiplicity, not to every stationary point of this rational function.