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) About Me  (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