\( 6 x^{2}-19 x+10 \) Select the correct choice below, and if necessary, fill in the answer box to complete your choice. A. \( 6 x^{2}-19 x+10=\square \) B. The trinomial is prime.

We begin with the trinomial \[ 6x^{2}-19x+10 \] and aim to factor it. Notice that the product of the coefficient of \(x^2\) and the constant term is \[ 6 \times 10 = 60. \] We need two numbers whose product is 60 and whose sum is \(-19\). The numbers \(-15\) and \(-4\) satisfy these conditions because \[ -15 \times -4 = 60 \quad \text{and} \quad -15 + (-4) = -19. \] Rewrite the middle term \(-19x\) using \(-15x\) and \(-4x\): \[ 6x^{2} - 15x - 4x + 10. \] Group the terms: \[ (6x^{2} - 15x) + (-4x + 10). \] Factor out the greatest common factor from each group: \[ 3x(2x - 5) - 2(2x - 5). \] Notice that \((2x-5)\) is common in both terms. Factor \((2x-5)\) out: \[ (2x-5)(3x-2). \] Thus, the factorization of the trinomial is \[ 6x^{2}-19x+10 = (2x-5)(3x-2). \] The correct choice is **A:** \[ 6x^2-19x+10=(2x-5)(3x-2). \]