Question

[1/(x^2+4x+3)]-[1/(x^2-3x-4)]

Answer 1

\[\frac{ 1 }{ x ^{2}+4x+3 } - \frac{ 1 }{ x ^{2}-3x-4 }\]

Answer 2

can you factor the denominators?

Answer 3

they factor to (x+3)(x+1) & (x-4)(x+1)
the common denominator is (x+3)(x+1)(x-4)

Answer 4

\[\frac{1}{x^{2}+4x+3}-\frac{1}{x^{2}-3x-4}\]
\[\frac{1}{(x+3)(x+1)}-\frac{1}{(x-4)(x+1)}\]
taking common
\[\frac{1}{(x+1)}[\frac{1}{(x+3)}-\frac{1}{(x-4)}]\]
now take LCM,

\[\frac{1}{(x+1)}[\frac{x-4-x-3}{(x+3)(x-4)}]\]

\[\frac{1}{(x+1)}[\frac{-7}{(x+3)(x-4)}]\]

\[\frac{-7}{(x+1)(x+3)(x-4)}\]