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what is quotient of (6-x)/(x^2 + 2x-3) divided by (x^2-4x-12/x^2 +4x +3? State restrictions. answer is -(x+1)/(x-1)(x+2) with restrictions of -3, -2,1,6. Don't understand why -3 and 6 are restrictions.
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when you divide by a fraction, you change the problem to multiply by its inverse (flip the fraction) the problem is \[ \frac{(6-x)}{x^2+2x-3} \cdot \frac{x^2+4x+3}{x^2-4x-12} \] after factoring we get \[ \frac{(6-x)}{(x+3)(x-1)} \cdot \frac{(x+3)(x+1)}{(x-6)(x+2)} \] it looks like we can cancel some terms. however, the simplified fraction does not match *exactly* this expression, because this expression is not defined at x= -3, 1, 6 or -2 to be complete, we make note of these numbers.
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