To rewrite the irrational square root of 63 as the product of an integer and another irrational number, we can start by simplifying \(\sqrt{63}\).
First, factor 63 into its prime factors: \[ 63 = 9 \times 7 = 3^2 \times 7 \]
Now, we can rewrite the square root of 63: \[ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3\sqrt{7} \]
Thus, we have expressed \(\sqrt{63}\) as the product of an integer (which is 3) and an irrational number (\(\sqrt{7}\)).
So, the final result in simplest form is: \[ \sqrt{63} = 3\sqrt{7} \]