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91Ó°ÊÓ

Use synthetic division to verify the upper and lower bounds of the real zeros of \(f\) \(f(x)=x^{3}+3 x^{2}-2 x+1\) (a) Upper: \(x=1\) (b) Lower: \(x=-4\)

Short Answer

Expert verified
Based on synthetic division, \(x=1\) is an upper bound for the real zeros of the given polynomial, while \(x=-4\) is not a lower bound.

Step by step solution

01

Test the Upper Bound

To test the upper bound, use synthetic division with \(x = 1\) as the divisor and the coefficients of the polynomial as the dividend. Set up the synthetic division table as follows:\n\n| 1 | 1 | 3 | -2 | 1 |\n|---------|-------|-----|------|----|\n| | 1 | 4 | 2 |\n\nAdd down the columns. The result will be the next row:\n\n| 1 | 1 | 4 | 2 |\n|----|-----|------|-----|\n| | 1 | 7 | 4 | 3 |\n\nThe last value in the final row, 3, is the remainder. Since this value is positive, we can conclude that \(x = 1\) is an upper bound for the real zeros.
02

Test the Lower Bound

To test the lower bound, use synthetic division with \(x = -4\) as the divisor. Set up the synthetic division table as follows:\n\n| -4 | 1 | 3 | -2 | 1 |\n|--------|-------|------|-----|-----|\n| | 1 | -1 | 2 |\n\nAdd down the columns. The result will be the next row:\n\n| -4 | 1 | -1 | 2 |\n|-----|-----| ------|-----|\n| | 1 | -3 | 10 | 9 |\n\nThe last value in the final row, 9, is the remainder. Since this value is positive, we can conclude that \(x = -4\) is not a lower bound for the real zeros.

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