Answer
No positive roots
$4$ or $2$ or $0$ negative roots
$4$ or $2$ or $0$ 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, $P(x)=x^4+x^3+x^2+x+12$, has no sign change, Therefore has no positive roots.
$P(-x)=x^4-x^3+x^2-x+12$, has $4$ sign changes, Therefore it has $4$ or $2$ or $0$ negative roots.
Thus, $P(x)$ has total of $4$ or $2$ or $0$ real roots.
