Inverting: zero at -2, pole at -5, HF gain 2

Required function: zero at $s=-2$, pole at $s=-5$, HF gain 2:

$$T(s) = 2\,\frac{s+2}{s+5} \quad(\text{DC gain } 0.8)$$

Inverting configuration: $Z_1$ from the input to the inverting input, $Z_2$ in feedback:

$$T(s) = -\frac{Z_2}{Z_1} = -\frac{Y_1}{Y_2}$$

With $Y_1 = \dfrac{1}{R_1} + sC_1$ (parallel $R_1C_1$) and $Y_2 = \dfrac{1}{R_2} + sC_2$ (parallel $R_2C_2$):

$$T(s) = -\frac{C_1}{C_2}\cdot\frac{s + \frac{1}{R_1C_1}}{s + \frac{1}{R_2C_2}}$$

              +---R2---+
              |        |
              +---||---+
       R1     |   C2   |
  +--/\/\--+  |        |
Vin        +--+-|-\    |
  +---||---+    |  >---+--- Vo
       C1   GND-|+/

Matching:

$$\begin{aligned} \frac{C_1}{C_2} &= 2 \\ \frac{1}{R_1C_1} &= 2 \\ \frac{1}{R_2C_2} &= 5 \end{aligned}$$

Normalized: $C_2 = 1$ F, $C_1 = 2$ F, $R_1 = \dfrac{1}{2\times2} = 0.25\ \Omega$, $R_2 = 0.2\ \Omega$.

Practical values (impedance scale $k_m = 10^5$, $C \to C/k_m$):

Check: DC gain $= R_2/R_1 = 0.8 = 2 \times 2/5$; HF gain $= C_1/C_2 = 2$.

Answer: $R_1 = 25\ \text{k}\Omega$, $C_1 = 20\ \mu$F, $R_2 = 20\ \text{k}\Omega$, $C_2 = 10\ \mu$F, realizing $-2(s+2)/(s+5)$. The minus sign is the 180° phase inversion of this configuration; a unity-gain inverter can follow if a positive sign is needed.