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Question
use the theorem on bounds to establish the best integral bounds for the roots of the following equation. 2x^3 - 5x^2 + 141 = 0 the best integral bounds are -4 < x < 2.
Step1: Identify the polynomial coefficients
The polynomial is $2x^{3}-5x^{2}+0x + 141$, with coefficients $a_3 = 2$, $a_2=-5$, $a_1 = 0$, $a_0=141$.
Step2: Apply the upper - bound theorem
We use synthetic division. First, test positive integers. Start with $x = 2$.
When we perform synthetic division with 2:
| 2 | 2 | -5 | 0 | 141 |
| - | - | - | - | - |
| 4 | - 2 | - 4 | ||
| 2 | - 1 | - 2 | 137 |
Since all the numbers in the bottom row (except the remainder) are non - negative when dividing by 2, 2 is an upper bound.
Step3: Apply the lower - bound theorem
Test negative integers. Start with $x=-4$.
When we perform synthetic division with - 4:
| -4 | 2 | -5 | 0 | 141 |
| - | - | - | - | - |
| - 8 | 52 | - 208 | ||
| 2 | - 13 | 52 | - 67 |
Since the numbers in the bottom row alternate in sign when dividing by - 4, - 4 is a lower bound.
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The best integral bounds are $-4