jmc

Setting $x=1,$ we get $$0=3P(1),$$so $P(x)$ has a factor of $x-1.$

Setting $x=-2,$ we get $$(-3)P(-1)=0,$$so $P(x)$ has a factor of $x+1.$

Setting $x=0,$ we get $$(-1)P(1)=2P(0).$$Since $P(1)=0,$ $P(0)=0,$ which means $P(0)$ has a factor of $x.$

Let $$P(x)=(x-1)(x+1)xQ(x).$$Then $$(x-1)x(x+2)(x+1)Q(x+1)=(x+2)(x-1)(x+1)xQ(x).$$This simplifies to $Q(x+1)=Q(x).$

Then $$Q(1)=Q(2)=Q(3)=Q(4)=\cdots .$$Since $Q(x)=Q(1)$ for infinitely many values of $x,$ $Q(x)$ must be a constant polynomial. Let $Q(x)=c,$ so $$P(x)=c(x-1)(x+1)x.$$Then $P(2)=6c$ and $P(3)=24c,$ so $$(6c{)}^2=24c.$$Solving, keeping in mind that $c\ne 0,$ we get $c=\frac{2}{3}.$ Then $P(x)=\frac{2}{3}(x-1)(x+1)x,$ and $$P\left(\frac{7}{2}\right)=\frac{2}{3}\cdot \frac{5}{2}\cdot \frac{9}{2}\cdot \frac{7}{2}=\boxed{\frac{105}{4}}.$$