Suppose $p(x)$ is a monic polynomial of degree $>0$ such that $P(0)<0$, which means $$p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ where $n\ge1,a_0,\cdots,a_{n-1}\in R$, and $a_0<0$.
It asks to prove that $S:=\{x\in R:p(x)<0 \}$ is bounded above.
The professor provided the answer sheet, however, I have a question with the first line. He set $$M:=(n-1)max\{-a_0,\cdots,-a_{n-1},1 \} $$ then shows M is an upper bound for S. Let $x>M$. Than for $k=0, \cdots, n-1$, $$x^k\ge x>M\ge -(n-1)a_k,$$ $$-\frac{a_k}{x^k}<\frac{1}{n-1},$$ $$\frac{a_k}{x^k}>-\frac{1}{n-1}.$$ Then, we can get $$p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \\=x^n(1+a_{n-1}x^{-1}+\cdots+a_1x^{-n+1}+a_0x^{-n}) \\>x^n(1+\frac{1}{n-1}+\cdots+\frac{1}{n-1} \\=0$$ Hence, $x\notin S$
I just want to know how the M get setted? I can't see a reason here.
In a monic polynomial, the coefficient of the highest power of $x$ is $1$, and therefore positive.
Now, when $x$ is large and positive, which of these is very large : $1,x,x^2,...,x^n$? Naturally, the higher the power of $x$, the bigger the value is. That is, after some time(that is, for very large $x$) , the highest power of $x$ starts taking over the rest of the other numbers. It is like this : when you add $10000$ and $1.5$ and $0.23$ and $0.001$, the smaller numbers don't contribute much to the sum of the four numbers, while of course $10000$ contributes the most. Think of the largest power of $x$ as being like $10000$, and the smaller powers as being like $0.23$,$1.5$ etc.
What is the coefficient of the largest power? It is $1$. So we have $x^n + ...$, where $...$ contains only smaller powers of $x$. If $x$ keeps on increasing, where will this polynomial go? Precisely where $x^n$ goes. Where does that go? Well, it goes to infinity. And that is why, you expect the polynomial to be positive after some time. What time?
Write down the polynomial: $$ P(x) = x^n + a_{n-1}x^{n-1} + ... + a_0 $$
You want this to be positive. What are the inequalities you get when it is made positive? $$ x^n + a_{n-1}x^{n-1} + ... + a_0 > 0 \iff x^n > -a_{n-1}x^{n-1} - ... - a_0 \\ \iff 1 > -a_{n-1}x^{-1} - ... - a_0x^{-n} $$
For $x > 0$. So, all you need, is a number $x$ which is large enough so that the right hand side is smaller than $1$.
One way of achieving this, is as follows: how many terms are on the right hand side? There are $n$ of them, namely $-a_{n-1}x^{-1},...,-a_0x^{-n}$(Your professor has made a mistake. This will be pointed out later on). Suppose we make each of these smaller than $\frac 1n$. Then the sum of $n$ terms, each smaller than $\frac 1n$, will definitely be smaller than $1$.
Pick any term $a_{n-k}x^{-k}$. You want it to be smaller than $\frac 1n$. First of all, make this simplification : since you want just an upper bound, you might as well assume that such a number is greater than $1$, in which case $x^{-1} \geq x^{-k}$ for all $k \geq 1$. Also, if any $a_k$ is positive or zero, then it will reduce the right hand side of the inequality, so we can ignore such $a_k$ in the calculation below, and assume that $a_k$ is positive, so that $-a_k > 0$.
Thus, $$ -a_{n-k}x^{-k} \leq -a_{n-k}^{-1} < \frac 1n \iff x > -n a_{n-k} $$
So, all we need is that $x $exceed the maximum of these quantities. That is what $M$ is : all you need to do is change the $n-1$ to $n$ above : that was the professor's mistake.