Solution
= Solution
Appending the node $-2$ to the existing <Newton interpolation polynomial> gives
$$
p_4(x)=p_3(x)+a_4x(x-1)(x-2)(x-3).
$$
At the new node,
$$
p_3(-2)=-82,
\qquad
(-2)(-3)(-4)(-5)=120.
$$
The condition $p_4(-2)=10$ therefore gives
$$
a_4=\frac{10-(-82)}{120}=\frac{23}{30}.
$$
Hence
$$
\boxed{
p_4(x)=4x-3x(x-1)+\frac73x(x-1)(x-2)
+\frac{23}{30}x(x-1)(x-2)(x-3)
}.
$$