Factorize (1) \( 9 x^{2}-4+(3 x-2)(x+3) \)

To factor the expression \( 9 x^{2}-4+(3 x-2)(x+3) \), we first simplify the component \( (3 x-2)(x+3) \): 1. \( (3 x-2)(x+3) = 3x^2 + 9x - 2x - 6 = 3x^2 + 7x - 6 \) Now, plug this back into the original expression: \[ 9 x^{2} - 4 + 3x^2 + 7x - 6 = (9x^2 + 3x^2 + 7x - 4 - 6) = 12x^2 + 7x - 10 \] Next, we can factor \( 12x^2 + 7x - 10 \). We need two numbers that multiply to \( 12 \times -10 = -120 \) and add up to \( 7 \). Those numbers are \( 15 \) and \( -8 \). Now we can rewrite the expression: \[ 12x^2 + 15x - 8x - 10 = (12x^2 + 15x) + (-8x - 10) \] Factor by grouping: \[ 3x(4x + 5) - 2(4x + 5) = (3x - 2)(4x + 5) \] So, the factored form of \( 9 x^{2}-4+(3 x-2)(x+3) \) is: \[ (3x - 2)(4x + 5) \]