^{4}+x³-x²-8

Its precalculus

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f(x)=2x^{4}+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.

Let's use the Rational Zeros Theorem:

±1, ±2, ±4, ±8

----------------------

±1, ±2

±1, ±2, ±4, ±8, ± 1/2

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

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

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!!

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