In the figure, AQPB and ASRC are squares, and AQS is an equilateral triangle. If QS  = 4 and  BC  = x, what is the value of x?
[asy]
unitsize(16);
pair A,B,C,P,Q,R,T;
A=(3.4641016151377544, 2);
B=(0, 0);
C=(6.928203230275509, 0);
P=(-1.9999999999999991, 3.464101615137755);
Q=(1.4641016151377544, 5.464101615137754);
R=(8.928203230275509, 3.4641016151377544);
T=(5.464101615137754, 5.464101615137754);
dot(A);dot(B);dot(C);dot(P);
dot(Q);dot(R);dot(T);
label("$A$", (3.4641016151377544, 2),E);
label("$B$", (0, 0),S);
label("$C$", (6.928203230275509, 0),S);
label("$P$", (-1.9999999999999991, 3.464101615137755), W);
label("$Q$", (1.4641016151377544, 5.464101615137754),N);
label("$R$", (8.928203230275509, 3.4641016151377544),E);
label("$S$", (5.464101615137754, 5.464101615137754),N);
draw(B--C--A--B);
draw(B--P--Q--A--B);
draw(A--C--R--T--A);
draw(Q--T--A--Q);
label("$x$", (3.4641016151377544, 0), S);
label("$4$", (Q+T)/2, N);[/asy]
1. **Identify the given information and the geometric properties:**
   - \( AQPB \) and \( ASRC \) are squares.
   - \( AQS \) is an equilateral triangle.
   - \( QS = 4 \).
   - We need to find \( BC = x \).

2. **Determine the side length of the squares:**
   - Since \( AQS \) is an equilateral triangle with \( QS = 4 \), each side of the equilateral triangle \( AQS \) is 4.
   - Therefore, the side length of the squares \( AQPB \) and \( ASRC \) is also 4.

3. **Calculate the coordinates of points \( A \), \( Q \), and \( S \):**
   - Place \( A \) at the origin \((0, 0)\).
   - Since \( AQS \) is an equilateral triangle with side length 4, and \( QS \) is horizontal, we can place \( Q \) at \((2, 2\sqrt{3})\) and \( S \) at \((4, 0)\).

4. **Determine the coordinates of points \( B \) and \( C \):**
   - Since \( AQPB \) is a square with side length 4, \( B \) is at \((0, -4)\).
   - Since \( ASRC \) is a square with side length 4, \( C \) is at \((4, -4)\).

5. **Calculate the length of \( BC \):**
   - The coordinates of \( B \) are \((0, -4)\) and the coordinates of \( C \) are \((4, -4)\).
   - The distance \( BC \) is the horizontal distance between \( B \) and \( C \), which is \( 4 - 0 = 4 \).

6. **Verify the calculation:**
   - The length of \( BC \) is indeed 4, as calculated from the coordinates.

The final answer is \( \boxed{4} \).