A Polynomial Problem..

Since $p(0) = 8$ , this implies that $p(x)$ is of the form $p(x) = ax(x-b)(x-c) + 8$ . And $p(3) = 8\implies p(x) = ax(x-3)(x-c) + 8$ . Now

$\begin{cases} p(2) = 2a(-1)(2-c) + 8 = 3 & \implies 4a - 2ac = 5 & ...(1) \\ p(4) = 4a(1)(4-c) + 8= 15 & \implies 16a - 4ac = 7 &...(2) \end{cases}$

$\begin{cases} 2 \times(1)-(2): & - 8a = 3 & \implies a = -\dfrac 38 \\ (1): & - \dfrac 32 + \dfrac 34 c = 5 & \implies c = \dfrac {26}3 \end{cases}$

Therefore $p(x) = \dfrac {x(x-3)(26-3x)}8+8 \implies p(x+4) = \dfrac {(x+4)(x+1)(14-3x)}8+8$ and the sum of coefficients of $p(x+4)$ is $p(1+4) = p(5) = \dfrac {5(2)(11)}8+8 = \boxed{21.75}$ .

$p(x)=A(x)(x-2)(x-3)(x-4)+x^{2}-1$

$p(x)$ is $3^{rd}$ -degree hence $A(x)$ is a constant

$p(x)=c(x-2)(x-3)(x-4)+x^{2}-1$

$p(0)=c(-2)(-3)(-4)-1=8$

$\Rightarrow c=-\frac{3}{8}$

Sum of the coefficients of $p(x+4)$ $=$ $p(5)$

$p(5)=-\frac{3}{8}(5-2)(5-3)(5-4)+5^{2}-1=21.75$