The order of polynomial $f(x)$ is $7$. If $f(x)+1$ can be divided by $(x-1)^4$, and $f(x)-1$ can be divided by $(x+4)^4$, can we know what $f(x)$ is?
I tried to let $f(x)+1=h(x)\,(x-1)^4$, $f(x)-1=g(x)\,(x+4)^4$, and supposed $$f(x)＝a_0+a_1x+\ldots+a_7x^7\,,$$ $$h(x）=h_0+h_1x+h_2x^2+h_3x^3\,,$$ $$g(x）=g_0+g_1x+g_2x^2+g_3x^3\,,$$ and use the Leibniz formula to compute the derivative of all orders of $f(x)$, and I get eight equations about the coefficients of $f(x)$, $h(x)$, and $g(x)$. It seems we can solve these equations to get $f(x)$, but this is too inconvenient. I want to ask if there is a better way to get $f(x)$.
We have
(1) $f(x)=(x-1)^4g(x)-1$ and
(2) $f(x)=(x+4)^4h(x)+1$
with some polynomial $g,h$ of degree $3$.
If we differentiate the equation (1), we see that $f'$ has at $1$ a zero of order $3$.
If we differentiate the equation (2), we see that $f'$ has at $-4$ a zero of order $3$.
Thus
$f'(x)=a(x-1)^3(x+4)^3$.
Can you proceed ?