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What is the maximum number of real zeros that [tex]P(x)[/tex] may have?

[tex]\[
P(x) = -26x^6 - 10x^4 - 27x^9 - 24x^{10} + 22x^2 - 9x^7 + 109
\][/tex]


Sagot :

To determine the maximum number of real zeros that the polynomial \( P(x) \) can have, we first need to look at its degree. The degree of a polynomial is the highest power of the variable \( x \) that appears in the polynomial with a non-zero coefficient.

Given the polynomial:
[tex]\[ P(x) = -26 x^6 - 10 x^4 - 27 x^9 - 24 x^{10} + 22 x^2 - 9 x^7 + 109 \][/tex]

Let's identify the terms and their degrees:

- \(-26 x^6\) has degree \(6\)
- \(-10 x^4\) has degree \(4\)
- \(-27 x^9\) has degree \(9\)
- \(-24 x^{10}\) has degree \(10\)
- \(22 x^2\) has degree \(2\)
- \(-9 x^7\) has degree \(7\)
- \(109\) (the constant term) has degree \(0\)

The highest degree among these terms is \(10\), which comes from the term \(-24 x^{10}\).

The Fundamental Theorem of Algebra states that a polynomial of degree \( n \) has exactly \( n \) roots in the complex number system (counting multiplicities). A real polynomial's roots can be real or complex, but the number of real roots cannot be greater than the degree of the polynomial.

For the given polynomial \( P(x) \), the highest degree is \(10\). Therefore, the maximum number of real zeros \( P(x) \) can have is 10.

Hence, the final result is:
[tex]\[ \boxed{10} \][/tex]