To find the product of \(4 \sqrt{12}\) and \(4 \sqrt{7}\), we can start by multiplying the two expressions:
\[ (4 \sqrt{12}) \cdot (4 \sqrt{7}) = 4 \cdot 4 \cdot \sqrt{12} \cdot \sqrt{7} \]
Calculating the constants:
\[ 4 \cdot 4 = 16 \]
Now we need to multiply the square roots:
\[ \sqrt{12} \cdot \sqrt{7} = \sqrt{12 \cdot 7} = \sqrt{84} \]
Therefore, we can rewrite the product as:
\[ 16 \sqrt{84} \]
Next, we simplify \(\sqrt{84}\). The expression \(84\) can be factored as follows:
\[ 84 = 4 \cdot 21 = 4 \cdot 3 \cdot 7 \]
This implies:
\[ \sqrt{84} = \sqrt{4 \cdot 21} = \sqrt{4} \cdot \sqrt{21} = 2 \sqrt{21} \]
Now substituting back into our product gives:
\[ 16 \sqrt{84} = 16 \cdot 2 \sqrt{21} = 32 \sqrt{21} \]
At this stage, our final result is:
\[ 32 \sqrt{21} \]
To determine whether \(32 \sqrt{21}\) is rational or irrational, we need to consider \(\sqrt{21}\). The number \(21\) is not a perfect square (it cannot be expressed as a square of an integer), meaning that \(\sqrt{21}\) is an irrational number.
Since the product of a rational number and an irrational number is always irrational, it follows that:
\[ 32 \sqrt{21} \]
is an irrational number.
In conclusion, the product of \(4 \sqrt{12}\) and \(4 \sqrt{7}\) in simplest form is:
\[ \boxed{32 \sqrt{21}} \]
And it is irrational.