What is (c^2+4c)/(c^2-4)*(c-2)/(3c+12) ?

The solution to (c^2+4c)/(c^2-4)*(c-2)/(3c+12) is c/(3(c+2))

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simplify (c^2+4c)/(c^2-4)*(c-2)/(3c+12)

simplify

\frac{c ^{2} + 4 c}{c ^{2} - 4} \cdot \frac{c - 2}{3 c + 12}

\frac{c}{3 ( c + 2 )}

Solution steps

\frac{c ^{2} + 4 c}{c ^{2} - 4} \cdot \frac{c - 2}{3 c + 12}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( c ^{2} + 4 c ) ( c - 2 )}{( c ^{2} - 4 ) ( 3 c + 12 )}

Factor

c^{2} - 4 : (c + 2) (c - 2)

= \frac{( c ^{2} + 4 c ) ( c - 2 )}{( c + 2 ) ( c - 2 ) ( 3 c + 12 )}

Cancel the common factor:

c - 2

= \frac{c ^{2} + 4 c}{( c + 2 ) ( 3 c + 12 )}

Factor

c^{2} + 4 c : c (c + 4)

= \frac{c ( c + 4 )}{( c + 2 ) ( 3 c + 12 )}

Factor

3 c + 12 : 3 (c + 4)

= \frac{c ( c + 4 )}{( c + 2 ) \cdot 3 ( c + 4 )}

Cancel the common factor:

c + 4

= \frac{c}{( c + 2 ) \cdot 3}

= \frac{c}{3 ( c + 2 )}

4 x^{2} + x + 5 = 0

4 v^{2} + 25 = - 20 v

\frac{3}{5} \cdot \frac{1}{2}

\frac{1}{6} + \frac{x + 2}{3} = 1

2 m - 1 > - 9

What is (c^2+4c)/(c^2-4)*(c-2)/(3c+12) ?
The solution to (c^2+4c)/(c^2-4)*(c-2)/(3c+12) is c/(3(c+2))