Answer
False.
Work Step by Step
Take $P(x)=(x+5)(x-1)=x^{2}+4x-5.$
We know the zeros are -5 and 2. Let's see if 2 is an upper bound (it should be):
$\left.\begin{array}{l}
2\lfloor \\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 &4&-5\\\hline
&2 &6\\\hline
1&6&6\end{array}$
All the entries in the bottom row are positive, so 2 is an upper bound.
But, $-2$ isn't a lower bound, because there is a zero $(x=-5)$ to the left of it.
This counterexample disproves the statement.
