What is the quotient \frac{6-x}{x^2+2x-3}\div\frac{x^2-4x-12}{x^2+4x+3} in simplified form?

Answer 1

To simplify this expression, we first need to rewrite the division as multiplication by the reciprocal of the second fraction:

\frac{6-x}{x^2+2x-3} \cdot \frac{x^2+4x+3}{x^2-4x-12}

Now, we can factor the denominators of the fractions to simplify further:

\frac{6-x}{(x+3)(x-1)} \cdot \frac{(x+3)(x+1)}{(x-6)(x+2)}

Next, we can cancel out common factors in the numerators and denominators:

\frac{6-x}{(x-1)} \cdot \frac{(x+1)}{(x+2)(x-6)}
=\frac{(6-x)(x+1)}{(x-1)(x+2)(x-6)}

Expanding the numerator gives:

\frac{6x + 6 - x^2 - x}{(x-1)(x+2)(x-6)}
=\frac{-x^2 + 5x + 6}{(x-1)(x+2)(x-6)}

So, the simplified form of the given expression is \frac{-x^2 + 5x + 6}{(x-1)(x+2)(x-6)}.