Answer
$1$ positive root and $2$ or $0$ negative roots
$3$ or $1$ roots
Work Step by Step
Descartes' Rule of Signs states that the possible number of the positive roots of a polynomial is equal to the number of sign changes in the coefficients or less than the number of sign changes by a multiple of $2$ AND the possible number of negative roots of a polynomial is equal to the number of sign changes or less than the total number of sign changes by a multiple of $2$ after substituting $−x$ for $x$. The substitution has the effect of negating all of the odd-power terms in the polynomial.
Therefore,
$P(x)=x^3-x^2-x-3$, has $1$ sign change, between $x^3$ and $-x^2$. Therefore, has a $1$ positive root.
$P(-x)=-x^3-x^2+x-3$, has $2$ sign changes, between $-x^2$ and $x$, and between $x$ and $-3$. Therefore, has $2$ or $0$ negative roots.
Therefore, It has a total of $3$ or $1$ real roots.
