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I have looked at the suggested related questions before asking.
From a First Course In Mathematical Analysis, David Brannan, page 148.Is the proof wrong?
zeros Localisation Theorem
Let $p(x) = {x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0}$,$x in R$,be a polynomial.
Then all zeros of p line in $( - M,M)$, where $M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|left| {{a_0}} right|} $

$r(x) = frac{{p(x)}}{{{x^n}}} - 1 = frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}},x in R - { 0} $ then, using the triangle inequality for $left| x right| > 1$

$left| {r(x)} right| = left| {frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}}} right| le left| {frac{{{a_{n - 1}}}}{x}} right| + ... + left| {frac{{{a_0}}}{{{x^n}}}} right|$

$ le max { left| {{a_{n - 1}}} right|,left| {{a_1}} right|,left| {{a_0}} right|}left( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}}} right)$

$ < Mleft( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}} + ...} right)$
$ = M$$frac{{{textstyle{1 over {left| x right|}}}}}{{1 - {textstyle{1 over {left| x right|}}}}} = frac{M}{{left| x right| - 1}}$it follows that if $ {x ge M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_0}} right|} } $ then $left| {r(x)} right| < 1$.
But surely, if $left| x right| = M$ then $left| {r(x)} right| < frac{M}{{left| x right| - 1}} = frac{M}{{M - 1}}$ is greater than 1!

Your error is that you wrote that the max is less than $M$ but you really should have just denoted it by $M-1$ to keep things tighter. In the end the numerator here should be $M-1$ if $M$ is defined the way you defined it. This is why I don't like the style of injecting this $1+$ into the definition of symbols, I would prefer to define $M$ to be the maximum and use functions of that if need be in the proof (and even in the statement, in this situation).

I understand now. Because $max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} le M - 1 < M$. So that
$max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} < M$ is true but now the tighter inequality is obliterated.
So looking back one line with $M$ in the sum causes a issue. $M - 1$ should have been carried through rather than $M$ to $frac{M}{{left| x right| - 1}}$. So a better bound on ${r(x)}$ is $frac{{M - 1}}{{left| x right| - 1}}$. I.e. $left| {r(x)} right| le frac{{M - 1}}{{left| x right| - 1}}$ then the argument from there is correct.