can u help me with this math question?

Question

[(c2 - 11c + 30)/(c3 + 6c2 + 9c)] · [(c2 - 3c)/(c2 - 25)] ÷ [(c2 - 9)/(c2 - 9c + 8)]

Answer 1

lets start by flipping the fraction on the end that we are dividing and then factoring completely everything else in the problem.
(c - 5)*(c -6)*c*(c + 3)*(c - 8)*(c - 1)
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c*(c+3)*(c+3)*(c+5)*(c-5)*(c-3)*(c+3)
next cancel things that are the same on the top and bottom.
whats left is
(c-6)(c-8)(c-1) / (c+3)^2 (c+5)(c-3)

Answer 2

[(c² - 11c + 30)/(c³ + 6c² + 9c)] • [(c² - 3c)/(c² - 25)] ÷ [(c² - 9)/(c² - 9c + 8)] =
{ [(c-5)(c-6)] / [c(c+3)²]} • { [c(c-3)]/[(c+5)(c-5)]} • { [(c-8)(c-1)] / [(c+5)(c-5)]} =
[ (c-6)(c-8)(c-1)] / [ (c-3)(c-5)(c+5)²]

not as much cancelled out as one might hope in such a problem.

Answer 3

factor out everything...

[(c-5)(c-6)/c(c+3)(c+3)] [c(c-3)/(c-5)(c=5)] [(c-1)(c-8)/(c-3)(c+3)]

cancel (c-5), (c), (c-3)

= (c-6)(c-1)(c-8)/(c+3)(c+3)(c+5)(c+3)
= (c-6)(c-1)(c-8)/(c+3)^3(c+5)