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Let the initial velocity of the bullet be $u$, the uniform deceleration inside the wall be $a$, and the distance penetrated be $s$.

According to the problem, after penetrating a distance of $s_1 = 2\text{ inches}$, the velocity of the bullet is halved, so $v_1 = \frac{u}{2}$.

Using the equation of motion $v^2 = u^2 - 2as$:
$\left(\frac{u}{2}\right)^2 = u^2 - 2a(2)$
$\frac{u^2}{4} = u^2 - 4a$
$4a = u^2 - \frac{u^2}{4} = \frac{3u^2}{4}$
$a = \frac{3u^2}{16}$

Let the bullet penetrate a further distance $x$ before coming to rest (final velocity $v = 0$).
Taking the velocity at the start of this second phase as $\frac{u}{2}$:
$0^2 = \left(\frac{u}{2}\right)^2 - 2ax$
$2ax = \frac{u^2}{4}$

Substituting the value of $a$:
$2 \left(\frac{3u^2}{16}\right) x = \frac{u^2}{4}$
$\frac{3u^2}{8} x = \frac{u^2}{4}$
$x = \frac{8}{4 \times 3} = \frac{2}{3}\text{ inches}$

Hence, the bullet will go a further $\frac{2}{3}\text{ inches}$ (or approximately $0.67\text{ inches}$) into the wall before its velocity becomes zero.