Answer
Lower bound = $-2$
Upper bound = $2.$
Work Step by Step
In trying to find rational zeros of $f(x)=x^{4}-3x^{2}-4$
(there are at most 4 real zeros),
we perform synthetic division and interpret the last row of the synthetic division table according to the Bounds on Zeros theorem.
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 4}{\pm 1}$
Descart's rule
$f(x)=x^{4}-3x^{2}-4$ has 1 change in signs$\Rightarrow$ 1 positive zero,
$f(-x)=x^{4}-3x^{2}-4$ has 1 change in signs$\Rightarrow$ 1 negative zero,
Testing (with synthetic division ) the highest possible $\displaystyle \frac{p}{q}=4$,
$\left.\begin{array}{l}
4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 0 & -3 & 0 &-4 \\\hline
& 4 & 16 & 52 & 208 \\\hline
1& 4 & 13 &52 & |\ \ 204 \end{array}$
- all the bottom entries are nonnegative,
$\Rightarrow$ 4 is an upper bound to the zeros of f but not a zero itself.
Now we search for lower upper bounds.
Testing (with synthetic division ) $\displaystyle \frac{p}{q}=2$,
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 0 & -3 & 0 &-4 \\\hline
& 2 & 4 & 2 & +4 \\\hline
1& 2 & 1 &2 & |\ \ 0 \end{array}$
- all the bottom entries are nonnegative, and 2 is a zero $\Rightarrow$ 2 is the upper bound.
(Being the only positive zero by Descartes' rule, we draw the same conclusion)
Since $f(x)=f(-x)$, the function is even and has a graph symmetric to the y-axis,
$\Rightarrow -2$ is a zero, and is the lower bound.
Lower bound = $-2$
Upper bound = $2.$
