Answer
$2$ or $0$ positive roots
No negative roots
$0$ is also a solution to an equation,
$3$ or $1$ possible real roots
Work Step by Step
Descartes' rule of signs: if a polynomial in one variable, $f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ...+ a_{1}x + a_{0}$, is arranged in the descending order of the exponents of the variable, then:
The number of positive real zeros of $f(x)$ is either equal to the number of sign changes in $f(x)$ or less than the number of sign changes by an even number.
The same rule applies to find the number of negative real zeros as well, but then we count the sign changes of $f(-x)$.
Thus, $P(x)=x^5+4x^3-x^2+6x$, has $2$ sign changes. Therefore has $2$ or $0$ positive roots.
$P(-x)=-x^5-4x^3-x^2-6x$, has no sign change, Therefore it has no negative roots.
Since $0$ is also a solution to an equation,
thus, $P(x)$ has total of $3$ or $1$ possible real roots.
