SOLUTION: Use the Theorem for bounds on zeros to find a bound on the real zeros of the polynomial function. f(x) = x^4 + 2x^2 - 3 a. -4 and 4 b. -3 and 3 c. -6 and 6 d

Upper Bound: 
If we divide a polynomial function f(x) by (x-r), where r is
POSITIVE using synthetic division, and we get all positive 
numbers on the bottom row of the synthetic division, then r is 
an upper bound to the real zeros of f(x).

 
Lower Bound: 
If we divide a polynomial function f(x) by (x-r), where r is
NEGATIVE, using synthetic division, and we get alternating 
signs for the numbers on the bottom row of the synthetic
division, then r is a lower bound to the real zeros of f(x). 

We try the smallest positive number among the choices,
which is 3

We use 3 to see if it is an upper bound to the zeros

3 |  1 0 2  0 -3
  |    3 6 24 72
     1 3 8 24 69   

All the numbers on the bottom row are positive, so 3 is
an upper bound for the zeros of f(x).

We try the negative number -3 that goes with that choice
to see if it is a lower bound:

-3 |  1  0  2   0 -3
   |    -3  9 -33 99
      1 -3 11 -33 96 

The signs on the bottom row of the synthetic division
alternate so -3 is a lower bound.

So all the zeros, if any, are between -3 and +3.

But, for that matter, since they are between -3 and +3,
they are also between -4 and +4.

They are also between -5 and +5

They are also between -6 and 6.

So actually every choice listed are also upper and lower
bounds.  Maybe they asked for the LEAST ones in absolute 
value that are on the list.  But if not, the author of
the book it came from has botched the problem. 

You might point out to your teacher that ALL of the 
choices given above are bounds for the zeros.

The only zeros are -1 and 1.  So any two numbers such that
the smaller number is less than -1 and the larger number is
greater than +1 are bounds for the zeroes.  This is really
a botched problem.  All the choices are correct!

Edwin