Find the degree, leading coefficients, and the maximum number of real zeros of the polynomial. \[ \begin{array}{l} f(x)=x^{4}-3 x^{2}+6-2 x^{5} \\ \text { Degree }=5 \end{array} \] Leading Coefficient \( = \) \( -2 \) Maximum number of real zeros \( =-2 x \) Question Help: Submit Question

The degree of the polynomial is 5, since the highest power of $x$ is $x^5$. The leading coefficient is $-2$, since this is the coefficient of the term with the highest power of $x$. To find the maximum number of real zeros, we can use Descartes' Rule of Signs. First, we look at the sign changes in the coefficients of the polynomial. Starting with the coefficient of $x^5$, we have a negative coefficient. Moving to the coefficient of $x^4$, we have a positive coefficient. There is one sign change. Continuing in this way, we have a negative coefficient for $x^3$, a positive coefficient for $x^2$, and a positive constant term. There are no more sign changes. Therefore, there is exactly one positive real zero. To find the number of negative real zeros, we can look at the polynomial $f(-x)$, which has the same number of negative real zeros as $f(x)$. Substituting $-x$ into the polynomial, we get \[f(-x) = (-x)^4 - 3(-x)^2 + 6 - 2(-x)^5 = x^4 - 3x^2 + 6 + 2x^5.\] This polynomial has two sign changes in its coefficients, so there are either two or zero negative real zeros. Therefore, the maximum number of real zeros is either one or three.