What is the quotient \frac{6-x}{x^2+2x-3}\div\frac{x^2-4x-12}{x^2+4x+3} in simplified form? State any restrictions on the variable.

1 answer

To divide two fractions, we invert the second fraction and multiply.

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

Inverting the second fraction:

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

Factor the polynomials:

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

Divide by multiplying the reciprocal:

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

Canceling out the common factors:

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

Simplify:

\frac{-x+6}{x+2}

Therefore, the quotient in simplified form is \frac{-x+6}{x+2}.

Restrictions on the variable are:
1. x cannot equal 1, -3, -2, or 6 because these values make the denominator(s) zero.