( x^2 + 5x +6)/( x^2 - 3x -10)
= (x+3)/(x-5) , x ≠ -2,5
suppose you had (2)(3)/[(3)(7)]
could you not cancel the equal factors to get 2/7
the above question follows the same concept, I canceled the factors (x+2).
x^2 + 5x +6/
x^2 - 3x -10
factor top and bottom
(x+2)(x+3) / [(x-5)(x+2]
= (x+3) / (x-5)
if x = 5 denominator is zero so exclude that point from the domain of x so I say A
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