91Ó°ÊÓ

Our polynomial is given as: \[p(x) = x^6 + 100x^2 + 5\] Notice that all the exponents are even in this polynomial. Even-degree terms always take positive values for both positive and negative values of x.

Let's analyze the values of each term when \(x > 0\), \(x < 0\), and \(x = 0\): 1) When \(x > 0\), - The term \(x^6\) is positive since any positive number raised to an even power remains positive. - The term \(100x^2\) is also positive since a positive number multiplied by \(100\) remains positive. - Lastly, the term \(5\) is always positive. 2) When \(x < 0\), - The term \(x^6\) is still positive, because a negative number raised to an even power becomes positive. - The term \(100x^2\) is positive, because squaring a negative number yields a positive result, and multiplying by \(100\) keeps it positive. - The term \(5\) remains positive as before. 3) When \(x = 0\), - The term \(x^6\) becomes \(0\). - The term \(100x^2\) becomes \(0\). - The term \(5\) remains positive.

Now, we analyze the value of the polynomial as the sum of its terms for each case mentioned previously: 1) When \(x > 0\), the polynomial can be written as: \[p(x) = (x^6) + (100x^2) + 5\] Since all terms are positive, their sum is also positive. Hence, \(p(x) > 0\) in this case. 2) When \(x < 0\), the polynomial can again be written as: \[p(x) = (x^6) + (100x^2) + 5\] As before, since all terms are positive, their sum is also positive. Therefore, \(p(x) > 0\) in this case as well. 3) When \(x = 0\), the polynomial can be written as: \[p(x) = (0) + (0) + 5\] In this case, \(p(x) = 5 > 0\).

Based on our analysis, it's clear that \(p(x) > 0\) for all real values of \(x\). As a result, a solution to the equation \(p(x) = 0\) does not exist among real numbers, and we can conclude that the polynomial \(p(x)\) has no real zeros.