Answer
- 14$y^{3}$ + 9x$y^{2}$ + 1
Work Step by Step
$\frac{-9x^{2}y^{3} - xy + 14xy^{4}}{-xy}$ =
RECALL:
$\frac{b^{n}}{b^{m}}$ = $b^{n-m}$
So, $\frac{-9x^{2}y^{3} - xy + 14xy^{4}}{-xy}$ = $\frac{-9x^{2}y^{3}}{-xy}$ + $\frac{-xy}{-xy}$ + $\frac{14xy^{4}}{-xy}$ =
9$x^{2-1}$$y^{3-1}$ + $x^{1-1}$$y^{1-1}$ - 14$x^{1-1}$$y^{4-1}$ =
9x$y^{2}$ + 1 - 14$y^{3}$ =
- 14$y^{3}$ + 9x$y^{2}$ + 1
