Answer
See the proof below.
Work Step by Step
As the order is 3, there should be at least one real zero $x=c$ to satisfy $P(c)=0$, which means that
$3c^3-c^2-6c+12=0$ divide both sides by 3, we have $c^3-\frac{c^2}{3}-2c+4=0$. For a rational zero
to satisfy this identify, $c$ must contain a factor $3$ to cancel the denominator. However, when we list
all the possible rational zeros ($\pm1,\pm1/3,\pm2,\pm2/3,\pm3,\pm4,\pm4/3,\pm6,\pm12$), there are
only 6 possible zeros ($\pm3,\pm6,\pm12$) contain factors of $3$. We test these zeros
$P(3)=3(3)^3-(3)^2-6(3)+12=66\ne0$,
$P(-3)=3(-3)^3-(-3)^2-6(-3)+12=-60\ne0$,
$P(6)=3(6)^3-(6)^2-6(6)+12=588\ne0$,
$P(-6)=3(-6)^3-(-6)^2-6(-6)+12=-636\ne0$,
$P(12)=3(12)^3-(12)^2-6(12)+12=4980\ne0$,
$P(-12)=3(-12)^3-(-12)^2-6(-12)+12=-5244\ne0$,
thus, we conclude that the polynomial does not have rational zeros.
Another way to prove there are no rational zeros is to graph the function as shown in the figure.
The solution $x=-1.874$ is not close to any of the possible rational zeros listed above.
