I am using the Descartes Rule of Signs.
f(x) = 3x^4 - 5x³ - x² - 8x + 4
positive zero: 2 or 0.
x = + - - +
There are 2 sign changes : + to - ; - to +.
negative zero: 1.
f(-x) = 3(-x^4) - 5(-x³) - (-x²) - 8(-x) + 4
f(-x) = 3x^4 + 5x³ - x² + 8x + 4
-x = + + - + +
there is only 1 change of sign. - to +.
There are 2 or 0 positive solutions and 1 negative solution.
Answer:
2
Step-by-step explanation:

Answer:

Step-by-step explanation:
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