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$\dfrac{4x^2-4}{x^2+4x+3} = \dfrac{4(x^2-1)}{(x+3)(x+1)}=\dfrac{4(x+1)(x-1)}{(x+3)(x+1)}$

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.