## College Algebra (10th Edition)

$\dfrac{8}{3x}; x \ne 0,2$
The given expression is equivalent to: $=\dfrac{4x-8}{-3x} \cdot \dfrac{12}{-6x+12}$ Factor each polynomial completely to obtain: $=\dfrac{2(2)(x-2)}{-3(x)} \cdot \dfrac{2(2)(3)}{-6(x-2)} \\=\dfrac{2(2)(x-2)}{-3(x)} \cdot \dfrac{2(2)(3)}{-2(3)(x-2)}$ Cancel the common factors to obtain: $\require{cancel} \\\\=\dfrac{2\cancel{(2)}\cancel{(x-2)}}{-3(x)} \cdot \dfrac{2(2)\cancel{(3)}}{-\cancel{2}\cancel{(3)}\cancel{(x-2)}} \\=\dfrac{2(2)(2)}{-3(x)(-1)} \\=\dfrac{8}{3x}; x \ne 0,2$