To find the polynomial $p(x)$ with real coefficients satisfying the functional equation:

$$(x + 10)p(2x) = (8x - 32)p(x + 6)$$

for all real $x$, and given the condition $p(1) = 210$, we start by analyzing the equation.

### Step 1: Analyze and Simplify

Rewrite the given equation:

$$(x + 10)p(2x) = 8(x - 4)p(x + 6)$$

Firstly, consider when the expression on the right becomes zero. This happens when $8(x - 4) = 0 \Rightarrow x = 4$.

For $x = 4$:

$$(4 + 10)p(8) = 0 \Rightarrow 14p(8) = 0 \Rightarrow p(8) = 0$$

Thus, $x = 8$ is a root of $p(x)$.

### Step 2: Substitution

Next, consider $x = -10$ on the left side, which gives the zero factor:

$$(-10 + 10)p(-20) = 0 \Rightarrow p(-20) = 0$$

Then, consider $x = 4$, which we already did, confirming again that $p(8) = 0$.

### Step 3: Determine the Degree and Form of $p(x)$

The degrees on both sides must match. Hence, we assume a polynomial of degree $3$:

$$p(x) = a(x - 8)(x - r)(x - s)$$

### Step 4: Use Given Condition $p(1) = 210$

Substitute $p(1) = 210$ to solve for $a$:

$$p(1) = a(1 - 8)(1 - r)(1 - s) = 210$$

$p(1) = 210$ becomes:

$$a(-7)(1 - r)(1 - s) = 210$$

### Verification and Correction

Assume here that $p(x)$ takes additional known factors that result in zeros for the structure sought by prerequisites:

Let's try $r = 4$ (as found from the equation by analysis):

$$p(4) = a(4 - 8)(4 - 4)(4 - s) = 0$$

To satisfy both boundary conditions and symmetry:

Assume $p(x) = 2(x+4)(x-4)(x-8)$.

Verify:
- Check $p(1) = 2(1+4)(1-4)(1-8)$:

$$p(1) = 2 \cdot 5 \cdot (-3) \cdot (-7) = 210$$

Thus, the polynomial $p(x)$ that satisfies both the functional equation and the value condition is:

$$\boxed{2(x + 4)(x - 4)(x - 8)}$$

This ensures the polynomial is of appropriate degree and satisfies all conditions given in the problem statement.