Answer:
[tex]\frac{1}{(x+3)(x+1)}[/tex]
Step-by-step explanation:
given [tex]\frac{x-2}{x^2+x-6}[/tex] ÷ [tex]\frac{x^2+5x+4}{x+4}[/tex]
factor the quadratic expressions in both fractions
x² + x - 6 = (x + 3)(x - 2) and x² + 5x + 4 = (x + 4)(x + 1)
Now leave the first fraction, change division to multiplication and turn the second fraction upside down.
= [tex]\frac{x-2}{(x-2)(x+3)}[/tex] × [tex]\frac{x+4}{(x+4)(x+1)}[/tex]
In the first fraction (x - 2) cancels and in the second fraction (x + 4) cancels, leaving
[tex]\frac{1}{(x+3)(x+1)}[/tex]
