1

a) For $a,\,b,\,c\in \mathbb{R}$ , let $f(x)=x^3+ax^2+bx+c$ and $M=\max\{1,|a|+|b|+|c|\}$. Show that $f(x)>0$ for $x>M$ and $f(x)<0$ for $x<-M$

b) Consider the following polynomial with integer coefficients $a_1,...,a_n$: $P(x)=x^n+a_1 x^{n-1}+...+a_n$. Show that every rational root of $P$ is an integer.

For the problem b) first I consider it is not true that $\frac{p}{q}:(p,q)=1$ is a root of this polynomial and putting this equation $P(x)=0$ and then contradict that $(p,q)\ne 1$. But what about a)?? Any help...

Bernard
  • 175,478
RAM_3R
  • 1,256

2 Answers2

1

Hint for a):

As $x>0$ if in the first case and $<0$ in the second case, all you have to prove is that $$\bigl|x\bigr|{}^3 >\bigl|ax^2+bx+c\bigr|,$$ and for that, by the triangle inequality, you can as well prove $$ x| ^3>|a||x|^2+|b||x|+|c|. $$ Observe that $$|x^3=|x||x|^2 >\bigl(|a|+|b|+|c|\bigr)|x|^2\quad\text{and}\quad |x|>1.$$

Bernard
  • 175,478
0

Not a rigorous proof: $x^3 > ax^2+ bx + c$ for some $x > r$

Consider the scenario where $r$ will be greatest. Then

$f(x) = x^3 - |a|x^2 - |b|x - |c|$

For $x = 1$, we get $f(1) = 1 - |a| - |b| - |c|$.

So if $1 \gt |a| + |b| + |c|$, then for all $x \ge 1$, $f(x) > 0$

and if $1 \lt |a| + |b| + |c|$, then for all $x \ge |a| + |b| + |c$, $f(x) > 0$

Likewise on the negative side.

sku
  • 2,643