Noise bound in FHE over the integers
Just expand it out.  You get $4\tau \cdot 2^\rho + 3 \cdot 2^\rho$.  Then, 

$\tau \cdot 4 \cdot 2^\rho = \tau \cdot 2^2 \cdot 2^\rho = \tau \cdot 2^{\rho + 2}$

and 

$3\cdot 2^\rho &lt; 4 \cdot 2^\rho = 2^{\rho+2}$.  

Put it back together, 

$\tau \cdot 2^{\rho + 2} + 2^{\rho + 2} &lt; \tau (2^{\rho + 2}+2^{\rho + 2}) = \tau 2^{\rho+3}$