MATH SOLVE

2 months ago

Q:
# What are the possible numbers of positive, negative, and complex zeros of f(x) = x6 + x5 + x4 + 4x3 − 12x2 + 12?

Accepted Solution

A:

Answer:Positive roots = 1 or 0Negative roots = 0, 2, or 4Complex roots = 0, 2, 4, or 6 Step-by-step explanation:We are given the polynomial:f(x) = -x⁶ - x⁵ - x⁴ - 4x³ − 12x² + 12 Now, by inspection, the highest degree is 6 and as such the polynomial is therefore a polynomial with more than 2 as a degree. Applying Descartes Rule of Signs to the polynomial : f(x) = -x⁶ - x⁵ - x⁴ - 4x³ − 12x² + 12 Signs are: - - - - - + There is only 1 sign change and thus, it means there is 1 or 0 positive roots To find number of negative roots, we will use f(-x);f(−x) = -(−x)⁶ - (−x)⁵ - (−x)⁴ - 4(−x)³ − 12(−x)² + 12(−x)f(-x) = -x⁶ + x⁵ - x⁴ + 4x³ − 12x² - 12 Signs are: - + - + - - There are 4 sign changes and thus, it means there are 4, 2 or 0 negative roots Thus; Complex roots = 0, 2, 4, or 6