[(c2 - c - 6)/(c2 - 4)] · [(c + 3)/(c2 + 6c + 9)] =?

Answer 1

c^2-c-6=(c-3)(c+2)
c^2+6c+9=(c+3)^2
c^2-4=(c+2)(c-2)
plugging in
(c-3)(c+2)/(c+2)(c-2)=c-3/c-2
(c+3)/(c+3)^2=1/(c+3)
the expression=(c-3)/(c-2)(c+3) answer

Answer 2

looks like this assignment is about factoring, and remembering the rules for multiplication

firs we need to factor all the quadratics and see what it looks like

(c2-c-6)=(c-3)(c+2)
(c2-4)=(c-2)(c+2)
(c2+6c+9)=(c+3)(c+3)

now combine the fractions through regular multiplication and get

[(c-3)(c+2)(c+3)] / [(c-2)(c+2)(c+3)(c+3)]

(c+3) and (c+2) in numerator and denominator cancels leaving:

(c-3) / (c-2)(c+3)

I did this pretty quick so you better check my algebra carefully

go over it and make sure you understand each step and you will be cranking these kind out without thinking in no time

Answer 3

first, factor the equations:

c2-c-6=(c-3)(c+2)
c2-4=(c+2)(c-2)
C2+6c+9=(c+3)(c+3)

then divide commen terms from the numerator and demoninators
c+2 goes away
C+3 goes away

(c-3)/((c-2)(c+3))
(c-3)/(c2+c-6)