\( \frac{y^{2}-2 y-15}{y^{2}-6 y+9} \div \frac{y-5}{y-3}=\square \) (Simplify your answer.)

To simplify the expression \( \frac{y^{2}-2 y-15}{y^{2}-6 y+9} \div \frac{y-5}{y-3} \), we first rewrite the division as multiplication by the reciprocal, giving us: \[ \frac{y^{2}-2 y-15}{y^{2}-6 y+9} \times \frac{y-3}{y-5} \] Next, we need to factor the polynomials. For the numerator \( y^{2}-2y-15 \): To factor this quadratic, we need two numbers that multiply to -15 and add to -2. Those numbers are -5 and 3, so we can factor it as: \[ (y-5)(y+3) \] For the denominator \( y^{2}-6y+9 \): This quadratic can be factored by noticing it is a perfect square: \[ (y-3)^2 \] Now we can substitute back into our expression: \[ \frac{(y-5)(y+3)}{(y-3)^2} \times \frac{y-3}{y-5} \] Next, we can cancel the common terms \( y - 5 \) and \( y - 3 \): After cancellation, we are left with: \[ \frac{y+3}{y-3} \] Thus, the simplified expression is: \[ \frac{y+3}{y-3} \]