## derivatives

Suppose $$f(x) = \frac{(x-3)^{4}}{x^{2}+2x}$$. Find the derivative an determine the values for which it is equal to 0. So $$f'(x) = \frac{x^{2}+2x(4(x-3)^{3}) - (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}}$$. But now how would I go about finding the values for which the derivative equals 0? $$f'(x) = \frac{x^{2}+2x(4(x-3)^{3}) (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} = 0$$. Is it possible to factor?

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 For any fraction, let's call it $\frac{A}{B}$, which part will make the whole thing equal to 0? A or B?
 A will yeah

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 Quote by plugpoint Suppose $$f(x) = \frac{(x-3)^{4}}{x^{2}+2x}$$. Find the derivative an determine the values for which it is equal to 0. So $$f'(x) = \frac{(x^{2}+2x)(4(x-3)^{3}) - (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}}$$. But now how would I go about finding the values for which the derivative equals 0? $$f'(x) = \frac{x^{2}+2x(4(x-3)^{3}) (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} = 0$$. Is it possible to factor? Thanks