Rationalize \(\cfrac{(4\sqrt{7} + \sqrt{2})}{(\sqrt{2} - 7)}\).

Rationalizing the denominator in a fraction means to eliminate the radical in the denominator. We do this by multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate of a binomial of the form \(a + b\) is \(a - b\). Hence, the conjugate of \(\sqrt{2}-7\) is \(\sqrt{2}+7\).

Here's how we rationalize the given fraction:

\[\cfrac{(4\sqrt{7} + \sqrt{2})}{(\sqrt{2} - 7)} \times \cfrac{(\sqrt{2} + 7)}{(\sqrt{2} + 7)}\]

The denominator becomes \((\sqrt{2} + 7)(\sqrt{2} - 7)\), which is a difference of squares and simplifies to \(2 - 49 = -47\).

The numerator becomes \((4\sqrt{7} + \sqrt{2})(\sqrt{2} + 7)\). Expand it to get \(-28\sqrt{7} - 6\sqrt{2} + 8\sqrt{14} + 14\). Notice that both terms in the numerator have a common factor of 2 which can be factored out to get \(2(-14\sqrt{7} - 3\sqrt{2} + 4\sqrt{14} + 7)\).

Divide the numerator and the denominator by \(-47\) to get \(-\cfrac{2(-14\sqrt{7} - 3\sqrt{2} + 4\sqrt{14} + 7)}{47}\).

When simplified, the expression becomes \(-\sqrt{14} - 6\), which corresponds to Option A.