Multiply. State any restrictions on the variable. \[ \frac{4 x^{2}+23 x+15}{16 x^{2}-9} \cdot \frac{4 x^{2}+9 x-9}{x^{2}+2 x-15} \] Simplify the rational expression. \( \frac{4 x^{2}+23 x+15}{16 x^{2}-9} \cdot \frac{4 x^{2}+9 x-9}{x^{2}+2 x-15}=\square \) (Simplify your answer.)

To multiply the given rational expressions, we start by factoring each part of the expressions where possible. 1. **Factoring the Expressions**: - The first numerator \( 4x^2 + 23x + 15 \) can be factored by finding two numbers that multiply to \( 4 \cdot 15 = 60 \) and add to \( 23 \). These numbers are \( 20 \) and \( 3 \): \[ 4x^2 + 23x + 15 = (4x + 3)(x + 5) \] - The first denominator \( 16x^2 - 9 \) is a difference of squares: \[ 16x^2 - 9 = (4x - 3)(4x + 3) \] - The second numerator \( 4x^2 + 9x - 9 \) factors to: \[ 4x^2 + 9x - 9 = (4x - 3)(x + 3) \] - The second denominator \( x^2 + 2x - 15 \) factors to: \[ x^2 + 2x - 15 = (x + 5)(x - 3) \] 2. **Setting Up the Expression**: Now, substituting back into the multiplication: \[ \frac{(4x + 3)(x + 5)}{(4x - 3)(4x + 3)} \cdot \frac{(4x - 3)(x + 3)}{(x + 5)(x - 3)} \] 3. **Cancelling Common Factors**: We can cancel out the common factors: - \( (4x + 3) \) in the numerator and denominator. - \( (4x - 3) \) in the numerator and denominator. - \( (x + 5) \) in the numerator and denominator. After cancellation, we get: \[ \frac{(x + 3)}{(x - 3)} \] 4. **Restrictions on the Variable**: The restrictions come from the denominators. The expression is undefined where any of the denominators are equal to zero: - \( 4x - 3 = 0 \) gives \( x \neq \frac{3}{4} \). - \( 4x + 3 = 0 \) gives \( x \neq -\frac{3}{4} \). - \( x + 5 = 0 \) gives \( x \neq -5 \). - \( x - 3 = 0 \) gives \( x \neq 3 \). Thus, the final simplified expression and restrictions are: \[ \frac{x + 3}{x - 3}, \quad x \neq 3, -5, -\frac{3}{4} \]