SOLUTION: P(x) = 2x^4 - 7x^3 + 5x^2 - 7x + 3 1. Find the upper bound and a lower bound for the real zeros of p(x). 2. Factor P(x) into first-degree factors. 3. Solve the equation where P

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Question 855560: P(x) = 2x^4 - 7x^3 + 5x^2 - 7x + 3
1. Find the upper bound and a lower bound for the real zeros of p(x).
2. Factor P(x) into first-degree factors.
3. Solve the equation where P(x) = 0.

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
P(x) = 2x^4 - 7x^3 + 5x^2 - 7x + 3
1. lower between 0 & 1, upper between 2 and 4. real zeroes: x = .5, 3
Using synthetic Division:
3 2 -7 5 -7 3
6 -3 6 -3
2 -1 2 -1 0
0.5 1 0 1
2 0 2 0
(x-3)(x-.5)(2x^2 + 2)
2. P(x) = 2(x-3)(x-.5)(x -i)(x+i)
3. P(x) = 2(x-3)(x-.5)(x -i)(x+i) = 0
x = .5, 3, -i, i


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