## Introductory Algebra for College Students (7th Edition)

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.