SOLUTION: Find the smallest positive integer and the largest negative integer that, by the Upper- and Lower-Bound Theorem, are upper and lower bounds for the real zeros of the polynomial fun
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Question 625973: Find the smallest positive integer and the largest negative integer that, by the Upper- and Lower-Bound Theorem, are upper and lower bounds for the real zeros of the polynomial function. (Enter your answers as a comma-separated list.)
P(x) = x3 + 6x2 − 3x − 8
x = ?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find the smallest positive integer and the largest negative integer that, by the Upper- and Lower-Bound Theorem, are upper and lower bounds for the real zeros of the polynomial function. (Enter your answers as a comma-separated list.)
P(x) = x3 + 6x2 − 3x − 8
**
Use rational roots theorem to solve:
...0...|.....1......6......-3......-8
...1...|.....1......7.......4.......-4 (Root exists between 1 & 2) (switch of signs)
...2...|.....1......8.......13......18 (2 is upper boundary, all numbers≥0
==================
.-1...|.....1......5.......-8........0 (-1 is a root)
P(x)=(x+1)(x^2+5x-8)
...0...|.....1......5.......-8
.-2...|.....1......3.......-14
.-3...|.....1......2.......-14
.-4...|.....1......1.......-12
.-5...|.....1......0.......-8
.-6...|.....1.....-1.......-2 (Root exists between -6 & -7)(switch of signs)
.-7...|.....1.....-2........6 (-7 is upper boundary, numbers alternate in sign)
upper bound: x=2
lower bound: x=-7
real roots: x=-1, two irrational roots between (1 & 2) and (-6 & -7)
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