Polynomial's division

When a polynomial $f(x)$ is divided by $(x-1)$ and $(x+5)$ , the remainders are $-6$ and $6$ respectively, so we get $f(1)=-6$ and $f(-5)=6$

Because $r(x)$ is the remainder when $f(x)$ is divided by $x^2+4x-5$ , degree of $r(x)$ less than $2$ .

Suppose that: $r(x)=ax+b$ , where $a,b$ are real numbers.

We have: $f(x)=(x^2+4x-5)g(x)+ax+b$

Or $\qquad \;f(x)=(x-1)(x+5)g(x)+ax+b$ .

If $x=1$ , then $a+b=f(1)=-6\qquad(1)$ .

If $x=-5$ , then $-5a+b=f(-5)=6\qquad(2)$ .

From $(1)$ and $(2)$ , we have: $a=-2; b=-4$ , implying $r(x)=-2x-4$ .

So, $r(-2)=-2.(-2)-4=\boxed{0}$

Write $r:=(x-1)(x+5),\quad a_1=-6,\ a_2=6, \  r_1=x-1,\ r_2=x+5\implies s_1=\frac{r}{r_1}=x+5,\ s_2=\frac{r}{r_2}=x-1$

then $(x+5)k_1\equiv1\mod(x-1)\Rightarrow k_1=\frac16$ $(x-1)k_2\equiv1\mod(x+5)\Rightarrow k_2=-\frac16$

$f(x)\equiv s_1k_1a_1+s_2k_2a_2\mod r\equiv (x+5)\frac16\cdot(-6)+(x-1)(-\frac16)6\equiv-\left[(x+5)+(x-1)\right]$

plug in $-2$ and get $0$