SOLUTION: The equation 2x^4 − 3x^3 − 14 x^2 − 22x − 8 = 0 has two real and two complex
solutions. What is the product of the two complex solutions?
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: The equation 2x^4 − 3x^3 − 14 x^2 − 22x − 8 = 0 has two real and two complex
solutions. What is the product of the two complex solutions?
Log On
Question 252309: The equation 2x^4 − 3x^3 − 14 x^2 − 22x − 8 = 0 has two real and two complex
solutions. What is the product of the two complex solutions? Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! Let's look at this using Descartes rule of signs. X = 4 and X = -2 will generate roots.
X = 4 divides the original polynomial with an answer of:
X = -1/2 divides the second polynomial with an answer of:
Set this polynomial = 0 and use quadratic. We get X =
(