## Precalculus (6th Edition)

$\color{blue}{-\dfrac{2}{x^4}}$
RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$ (4) $(ab)^m = a^mb^m$ (5) $(a^m)^n=a^{mn}$ (6) $a^m \cdot a^n = a^{m+n}$ (7) $a^0=1, a\ne0$ Use rule (6) above to obtain: $=\dfrac{-8xy^{1+3}}{4x^5y^4} \\=\dfrac{-8xy^4}{4x^5y^4}$ Divide the coefficients by cancelling out the common factors to obtain: $\require{cancel} \\=\dfrac{-\cancel{8}^2xy^4}{\cancel{4}x^5y^4} \\=\dfrac{-2xy^4}{x^5y^4}$ Use rule (2) above to obtain: $=-2x^{1-5}y^{4-4} \\=-2x^{-4}y^{0}$ Use rule (7) above to obtain: $=-2x^{-4}(1) \\=-2x^{-4}$ Use rule (1) above to obtain: $=-2 \cdot \dfrac{1}{x^4} \\=\dfrac{-2}{x^4} \\=\color{blue}{-\dfrac{2}{x^4}}$