Question: I would like to solve: 

$M =ABC$ for $B$ where
$M$, $A$, and $C$ are known rectangular matrices of compatible dimensions.

e.g. $(1000\times 200) = (1000\times 30)\cdot B\cdot (14000\times 200)$

I am familiar with the process for solving for $B$ when there is no term $C$.

Thank you.

Answer: 
In this situation, f decreases through its zero. Assuming convexity, you can show geometrically that Newton will undershoot if you start left of the zero and overshoot if you start right of the zero (and left of the minimum, which is not guaranteed by your assumptions). This is because f lies above its tangent lines. As a result you have at most one step to get to the left of the zero, after which you have a bounded increasing sequence.
f is differentiable and convex , $f(c)=0$ .If $A(x\_0,f(x\_0))$ is a point of $C\_f$ with $x\_0 <c$ then the tangent at $x=x\_0$ is $(e):y-f(x\_0)=f'(x\_0)(x-x\_0)$
for $y=0:$ $$x-x\_0= -\frac{f(x\_0)}{f'(x\_0)}$$
and because $f$ is convex then $(e)$ is bellow $C\_f$

So $x\_0<x<c$

For the sequence we have ${x\_0}=a<c$ as a result of applying the above $x\_0<x\_1<c$
with induction we get $x\_0<x\_1<x\_2<...<x\_n$
So the sequence converges.
So $lim{n\to +\infty}({x\_{n+1}-x\_n})=0\Rightarrow lim{n\to +\infty}(-\frac{f(x\_n)}{f'(x\_n)})=0 $ 

We know that $f'$ is continious because it is increasing and not zero $\forall x\leq c$

So $ lim{n\to +\infty} (f(x\_n))=0$
so $x\_n\rightarrow c$