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use synthetic division to verify the upper and lower bounds of the zeroes of f(x)=2x4+x³-x²-8

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f(x)=2x4+x³-x²-8
f(2)=32+8-4-8=28
f(-2)=32-8-4-8=12
 
-2|  2   1   -1     0    -8
          -4    6  -10   20
      2   -3   5  -10   12  There are no zeroes less than -2
 
2| 2  1  -1    0   -8
        4  10  18  36
    2  5    9  18  28        There are no zeroes greater than 2
 
The way I chose 2 and  -2 was by drawing a picture. 
 
f(x) = 2x^4 + x^3 - x^2 - 8

Let's use the Rational Zeros Theorem: "The possible rational zeros (if any) are positive or negative fractions whose numerators are factors of 8 and whose denominators are factors of 2" to make a list of possible zeros

  ±1, ±2, ±4, ±8
----------------------
        ±1, ±2

 ±1, ±2, ±4, ±8, ± 1/2

"If P(x) is a polynomial and we know that P(a) >0 and P(b) < 0, then somewhere between "a" and "b" is a zero of P(x)."

Let's analyze ±1 and ±2 from the list:

f(-2) = 32 - 8 - 4 - 8 = 12 > 0  
f(-1) = 2 - 1 - 1 - 8 = - 8 < 0    so, there is, at least, one zero between "- 2" and "- 1"

f(1) =  2 + 1 - 1 - 8 = - 6 < 0   It looks like there is no any zero between "- 1" and "1"
Note, that f(0) = - 8 < 0

f(2) = 32 + 8 - 4 - 8 = 28 > 0   so, there is, at least, one zero between "1" and "2" 

Looks like we bound on zeros. It's an interval: (- 2, 2) . The lower bound is "- 2" and upper bound "2"

Now, let's use synthetic division to prove it. 

2 > 0, if all numbers on the bottom will be positive or zero, then 2 is a upper bound of zeros on f(x).

 2 |   2    1   - 1      0     - 8
    |        5    12    22      44   
        2    6    11    22     36      <---- all numbers are positive!!

- 2 < 0, if all number on the bottom will be alternating in sign, then "- 2" is a lower bound of zeros on f(x)

 - 2 |  2     1  - 1       0    - 8  
      |      - 4     6   - 10    20   
         2   - 3    5    - 10    12     <----- all numbers are alternating in sign!!