The question is consider the polynomial $P(X) = X^d + a_{d−1}X^{d−1} + · · · + a_1 X + a_0$, where $d \geq 1$ and $a_0$, . . . , $a_{d−1}$ are real numbers. Define the number $M := 1 + |a_{d−1}| + · · · + |a_1| + |a_0|$. Prove the following:
(1) For every $x \in \mathbb{R}$ such that $|x| \geq M$, $$ \left| \frac{a_{d−1}}{x} + \frac{a_{d−2}}{x^2} + · · · + \frac{a_{1}}{x^{d-1}}+\frac{a_{0}}{x^d} \right| <1. $$ (2) For every $x \in \mathbb{R}$ such that $|x| \geq M$, $P(x)= x^d (1 + e)$ for some $e \in \mathbb{R}$ such that $|e| < 1$.
(3) For every $x \in \mathbb{R}$ such that $|x| \geq M$, $P(x)$ is not equal to $0$.
I honestly do not know where to start, I tried to manipulate the absolute values in the equation but can't seem to get anywhere, and I think its because I don't really know the point of this proof or what it is exactly I am proving. Can anyone help me out? Thanks!
For the first part, you can major the expression substituting x with M (x^2>x, ... , x^d>x so you can get rid of the exponents) which will give something smaller than 1 because M > a_0+...+a_(d-1).
For the second part just put x^d in evidence in the formulation of P(x). You'll get x^d(1 + (...)). Because you've proven a) already the result is true.
As for the third part, notice that P(x)=0 only if x=0 or e=1 (using 2)). Both are Impossible from M>1 and e<1