I'm an engineer, not a math guy. Please use small words if possible. ;-)
I am going through this neat paper on a method of partial fraction decomposition by repeated synthetic division. On page 157 it is covering case with repeated irreducible quadratics in denominator - and it says this:
"Multiply both sides by the denominator...then We reduce the right hand side modulo $x^2+bx+c $ by sending it to the field $\mathbb R[x]/(x^2+bx+c) = $ {$px+q|p,q\in \mathbb{R},x^2+bx+c=0$}. Modulo $x^2+bx+c$....
I have no clue how to "reduce the r.h.s. modulo by sending it to the field". Can someone give me example of how to do such? Preferably with real equations i can ponder like an engineer.
UPDATE 1: Adding real example we can use. Let, $$\frac{N(s)}{D(s)}=\frac{10(s+1)}{(s+1)^3(s^2+5s+6)^3}=\frac{K_1}{s+1}+\frac{K_2}{(s+1)^2}+\frac{K_3}{(s+1)^3}+\frac{M_1s+C_1}{s^2+5s+6}+\frac{M_2s+C_2}{(s^2+5s+6)^2}+\frac{M_3s+C_3}{(s^2+5s+6)^3}$$
Multiplying both sides by D(s) I get, $$N(s) = 10(s+3) = K_1(s+1)^2(s^2+5s+6)^3+K_2(s+1)(s^2+5s+6)^3+K_3(s^2+5s+6)^3+(M_1s+C_1)(s+1)^3(s^2+5s+6)^2+(M_2s+C_2)(s+1)^3(s^2+5s+6)+(M_3s+C_3)(s+1)^3$$
How do i proceed now?
$$\begin{align} 10(s+3) &= K_1(s+1)^2(s^2+5s+6)^3 \\ & +K_2(s+1)(s^2+5s+6)^3 \\ &+ K_3(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^3 \end{align}$$
Let $s = -1$ and you get $20 = 8K_3$. So $K_3 = \dfrac 52$
$$\begin{align} 10(s+3) &= K_1(s+1)^2(s^2+5s+6)^3 \\ & +K_2(s+1)(s^2+5s+6)^3 \\ &-\dfrac 52(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^3 \end{align}$$
$$\begin{align} -\dfrac 52 (s + 3) (s + 1) (s^4 + 11 s^3 + 46 s^2 + 88 s + 68) &= K_1(s+1)^2(s^2+5s+6)^3 \\ & +K_2(s+1)(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^3 \end{align}$$
$$\begin{align} -\dfrac 52 (s + 3)(s^4 + 11 s^3 + 46 s^2 + 88 s + 68) &= K_1(s+1)(s^2+5s+6)^3 \\ & +K_2(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^2(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^2(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^2 \end{align}$$
Let $s=-1$ again and you get...
Added 11/22/2020
Your original equation is wrong. It should look like
\begin{align} \frac{10}{(s+1)^2 (s+2)^3 (s+3)^3} &=\\ &\frac{K_1}{(s+1)} + \frac{K_2}{(s+1)^2} + \\ &\frac{L_1}{(s+2)} + \frac{L_2}{(s+2)^2} + \frac{L_3}{(s+2)^3} + \\ &\frac{M_1}{(s+3)} + \frac{M_2}{(s+3)^2} + \frac{M_3}{(s+3)^3} \end{align}
You then get the equation
\begin{align} 10 &= K_1(s+1)(s+2)^2 (s+3)^3 \\ &+ K_2(s+2)^3 (s+3)^3 + \\ &+ L_1(s+1)^2 (s+2)^2 (s+3)^3 \\ &+ L_2(s+1)^2 (s+2) (s+3)^3 \\ &+ L_3(s+1)^2 (s+3)^3 \\ &+ M_1 (s+1)^2 (s+2)^3 (s+3)^2 \\ &+ M_2 (s+1)^2 (s+2)^3 (s+3) \\ &+ M_3 (s+1)^2 (s+2)^3 \\ \end{align}
Now let $s=-1$. You get $10=8K_2 \implies K_2 = \dfrac 54$. Simplifying, we get
\begin{align} \dfrac 58 (s + 1) (9 s^5 + 124 s^4 + 685 s^3 + 1902 s^2 + 2668 s + 1528) &= K_1(s+2)^3 (s+3)^3 \\ &+ L_1(s+1) (s+2)^2 (s+3)^3 \\ &+ L_2(s+1) (s+2) (s+3)^3 \\ &+ L_3(s+1) (s+3)^3 \\ &+ M_1 (s+1) (s+2)^3 (s+3)^2 \\ &+ M_2 (s+1) (s+2)^3 (s+3) \\ &+ M_3 (s+1) (s+2)^3 \\ \end{align}
Again let $s=-1$. You get $-45 = 8K_1 \implies K_1 = -\dfrac{45}{8}$. Simplifying, we get
\begin{align} -\dfrac 54 (4 + s)(52 + 70 s + 39 s^2 + 10 s^3 + s^4) &= L_1 (s+2)^2 (s+3)^3 \\ &+ L_2 (s+2) (s+3)^3 \\ &+ L_3 (s+3)^3 \\ &+ M_1 (s+2)^3 (s+3)^2 \\ &+ M_2 (s+2)^3 (s+3) \\ &+ M_3 (s+2)^3 \\ \end{align}