+6248 \(\dfrac{4x^2-4}{x^2+4x+3} = \dfrac{4(x^2-1)}{(x+3)(x+1)}=\dfrac{4(x+1)(x-1)}{(x+3)(x+1)}\)
True
True
This is a bit tricky. Clearly there are two terms in both the numerator and denominator as written but we should cancel out the (x+1) term in both the numerator and denominator and obtain
\(\dfrac{4(x-1)}{x+3}\)
This has only a single term in both the numerator and denominator.
The answer your teacher wants to see depends on them.
