Simplify the expression. \( \frac{(2 x+7)(3 x+4)}{6 x^{2}+13 x-28} \) (A) 1 (B) \( \frac{3 x+4}{3 x-4} \) (C) \( \frac{1}{13 x} \) (D) \( \frac{29}{13} \)

To simplify the expression \( \frac{(2 x+7)(3 x+4)}{6 x^{2}+13 x-28} \), we first need to factor the denominator \( 6 x^2 + 13 x - 28 \). 1. We look for two numbers that multiply to \( 6 \times (-28) = -168 \) and add to \( 13 \). The numbers \( 21 \) and \( -8 \) work because \( 21 - 8 = 13 \) and \( 21 \cdot (-8) = -168 \). 2. Rewrite the middle term: \[ 6x^2 + 21x - 8x - 28 \] 3. Now, factor by grouping: \[ 3x(2x + 7) - 4(2x + 7) = (3x - 4)(2x + 7) \] 4. So, the expression now looks like: \[ \frac{(2x + 7)(3x + 4)}{(3x - 4)(2x + 7)} \] 5. We can cancel out \( (2x + 7) \) from the numerator and denominator (provided \( 2x + 7 \neq 0 \)): \[ = \frac{3x + 4}{3x - 4} \] Thus, the simplified expression is \( \frac{3x+4}{3x-4} \), which corresponds to option (B).