The function $f$ is defined, for all values of $x$, by $f(x)=(x+3)(1-2x)^2$. Find the range of values of $x$ for which $f$ is a decreasing function.

1. Find the derivative of $f(x)$

Using the product rule and chain rule:
$f'(x) = 1 \cdot (1-2x)^2 + (x+3) \cdot 2(1-2x)(-2)$
$f'(x) = (1-2x)^2 - 4(x+3)(1-2x)$
$f'(x) = (1-2x)[(1-2x) - 4(x+3)]$
$f'(x) = (1-2x)(-6x - 11)$

2. Set up the inequality for a decreasing function

For $f(x)$ to be a decreasing function, $f'(x) < 0$.
$(1-2x)(-6x - 11) < 0$
$(2x - 1)(6x + 11) < 0$

3. Solve the quadratic inequality

The roots of the quadratic equation $(2x - 1)(6x + 11) = 0$ are $x = \frac{1}{2}$ and $x = -\frac{11}{6}$.
Since the coefficient of $x^2$ is positive, the parabola opens upwards.
The inequality $(2x - 1)(6x + 11) < 0$ is satisfied between the roots:
$-\frac{11}{6} < x < \frac{1}{2}$