5. Let the polynomial $p(x)$ have a degree not exceeding 3, and $p(0)=$ $1, p(3)=0,|p(2+x)|=|p(2-x)|$. If the leading coefficient of $p(x)$ is negative, then $p(x)=$ $\qquad$ $ـ$
5. $p(x)=-\frac{1}{6}(x-1)(x-2)(x-3)$.

From $p(3)=0$ we know that $(x-3)$ is a factor of $p(x)$, and from $|p(2+x)|=|p(2-x)|$ we know that $|p(3)|=|p(1)|=0$, so $(x-1)$ is also a factor of $p(x)$.

Let $p(x)=(x-1)(x-3)(a x+b)$, and from $p(0)=1$ we get $f(0)=(-1)(-3) b=1, b=\frac{1}{3}$, and from $|p(2+x)|=|p(2-x)|$ we get
$\left.\left\lvert\, a x+ 2 a+\frac{1}{3}\right.\right) \left.|=|-a x-\left(2 a+\frac{1}{3}\right) \right\rvert\,$,
letting $x=2$ we get, $\left|4 a+\frac{j}{3}\right|=\left|\frac{1}{3}\right|$.
Thus, $a=0$ or $a=-\frac{1}{6}$, accordingly.