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

1) Let's evaluate the domain of

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

For the expression

\frac{c^{2} + 4 c}{c^{2} - 4}

to exist, you must check:

c^{2} - 4 \neq 0

For the expression

\frac{c - 2}{3 c + 12}

to exist, you must check:

3 c + 12 \neq 0

2) Let's solve in

ℝ

c^{2} - 4 \neq 0

The two numbers whose square is

4

are

-2

and

2

Solution found:

c \in ℝ \ {-2, 2}

3) Let's solve in

ℝ

3 c + 12 \neq 0

ⓘ We subtract

12

from both sides

3 c \neq -12

ⓘ Both sides are divided by

3

c \neq - \frac{12}{3}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

3

, their greatest common divisor:

- \frac{12}{3}

-4

Solution found:

c \neq -4

Let's build the table of the different terms to determine the global solution:

▶Domain of definition:

c \in ℝ \ {-4, -2, 2}