#### Answer

Makes sense.

#### Work Step by Step

Try an example: $2x^{3}-4x^{2}+7x-14$
$\left[\begin{array}{lll}
(2x^{3}-4x^{2})+(7x-14) & ... & (2x^{3}+7x)+[-4x^{2}-14] \\
=2x^{2}(x-2)+7(x-2) & & =x(2x^{2}+7)-2(2x^{2}+7)\\
=(x-2)(2x^{2}+7) & & =(2x^{2}+7)(x-2)
\end{array}\right]$
which are equivalent, since multiplication is commutative.
Two different groupings, same factorizations.
Makes sense.