Question

Please show steps and show restrictions

Answer 1

$\dfrac{x^2+7x+12}{x^2-x-12}\\\\=\dfrac{x^2 +3x+4x +12}{x^2 -4x+3x-12}\\\\=\dfrac{x(x+3)+4(x+3)}{x(x-4)+3(x-4)}\\\\=\dfrac{(x+4)(x+3)}{(x-4)(x+3)}\\\\=\dfrac{x+4}{x-4}\\\\\text{Restrictions:}~ x\neq 4 ,~ x\neq -3$