Question

what is the maximum number of real zeros that p(x) may have?
p(x) = -23x⁸ + 13x⁵ - 9x⁷ - 25x⁹ - 18x⁶ + 26x³ + 139
max number of real zeros

Explanation:

Step1: Identify the degree of the polynomial

The degree of a polynomial is the highest power of \( x \) in the polynomial. For \( P(x)= -23x^{8}+13x^{5}-9x^{7}-25x^{9}-18x^{6}+26x^{3}+139 \), we look at the exponents of \( x \): 8, 5, 7, 9, 6, 3. The highest exponent is 9. So the degree of \( P(x) \) is 9.

Step2: Recall the Fundamental Theorem of Algebra (for real zeros)

The Fundamental Theorem of Algebra states that a polynomial of degree \( n \) has at most \( n \) real zeros (counting multiplicities, but for the maximum number of distinct real zeros, it's also at most \( n \)). Since our polynomial has degree 9, the maximum number of real zeros it can have is equal to its degree.

Answer:

9