\( \begin{aligned} \text { 4. } \frac{x^{2}+2 x-3}{3 x+3} \div \frac{x^{2}+5 x+6}{2 x+2} & =\square \\ & =\square\end{aligned} \)

To solve the expression \( \frac{x^{2}+2x-3}{3x+3} \div \frac{x^{2}+5x+6}{2x+2} \), we can rewrite it as a multiplication by the reciprocal: \[ \frac{x^{2}+2x-3}{3x+3} \times \frac{2x+2}{x^{2}+5x+6} \] Now, let's factor the polynomials: 1. Factor \( x^{2}+2x-3 = (x+3)(x-1) \) 2. Factor \( 3x+3 = 3(x+1) \) 3. Factor \( x^{2}+5x+6 = (x+2)(x+3) \) 4. Factor \( 2x+2 = 2(x+1) \) Now substituting these factored forms into the expression gives us: \[ \frac{(x+3)(x-1)}{3(x+1)} \times \frac{2(x+1)}{(x+2)(x+3)} \] Next, we can cancel out common factors in the numerator and denominator: - \( (x+3) \) in the numerator and denominator, - \( (x+1) \) in the numerator and denominator. This simplifies to: \[ \frac{2(x-1)}{3(x+2)} \] Now we have our final simplified expression: \[ \frac{2(x-1)}{3(x+2)} = \square \] This means: 1. The simplified form of the division is \( \frac{2(x-1)}{3(x+2)} \). 2. If you need a specific value, you can substitute any valid \( x \) (except where the original denominator equals zero) to find out the corresponding value.