#1**+1 **

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

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.

Rom Oct 19, 2018