Control theory - how to represent these transfer functions?
first thing to do is divide both numerator and denominator by \$s\$ or \$s^2\$, whatever power of \$s\$ so that no positive powers exist in the denominator.  (we want integrators, \$s^{-1}\$, in the blocks, not differentiators.)  then your transfer function will appear like

$$ H(s) = \frac{Y(s)}{X(s)} = \frac{b_0 + b_1 s^{-1} + b_2 s^{-2}... + b_N s^{-N}}{a_0 + a_1 s^{-1} + a_2 s^{-2}... + a_N s^{-N}} $$

then divide both numerator and denominator by the constant term in the denominator, \$a_0\$, so that now your transfer function will look like

$$ H(s) = \frac{Y(s)}{X(s)} = \frac{b_0 + b_1 s^{-1} + b_2 s^{-2}... + b_N s^{-N}}{1 + a_1 s^{-1} + a_2 s^{-2}... + a_N s^{-N}} $$

(the other \$a_n\$ and \$b_n\$ are different now than they were before because they all were divided by \$a_0\$.)

now rearrange things:

$$ Y(s) \left(1 + a_1 s^{-1} + a_2 s^{-2}... + a_N s^{-N} \right) = X(s) \left(b_0 + b_1 s^{-1} + b_2 s^{-2}... + b_N s^{-N} \right) $$

...

$$ Y(s) + a_1 s^{-1}Y(s) + a_2 s^{-2}Y(s)... + a_N s^{-N}Y(s) = \\ \quad  b_0 X(s) + b_1 s^{-1}X(s) + b_2 s^{-2}X(s)... + b_N s^{-N}X(s) $$

...

$$ Y(s) = b_0 X(s) + (b_1 (s^{-1}X(s)) + b_2 (s^{-2}X(s))... + b_N (s^{-N}X(s)) \\ - a_1 (s^{-1}Y(s)) - a_2 (s^{-2}Y(s))... - a_N (s^{-N}Y(s)) $$

that defines \$Y(s)\$ in terms of \$X(s)\$, integrated versions of \$X(s)\$ and integrated versions of \$Y(s)\$, the latter is feedback from the output.

try drawing a block diagram out of that.