Factor the trinomial. \( 20 a^{2}-23 a d+6 d^{2} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( 20 a^{2}-23 a d+6 d^{2}=\square \) (Factor completely.) B. The trinomial is prime.

To factor the trinomial \( 20a^{2} - 23ad + 6d^{2} \), we'll look for two binomials of the form \((ma + nd)(pa + qd)\) such that: \[ (mp)a^{2} + (mq + np)ad + (nq)d^{2} = 20a^{2} - 23ad + 6d^{2} \] **Step 1: Identify Possible Pairs** 1. **For \(a^{2}\) term:** \(mp = 20\) Possible pairs: \((5, 4)\), \((10, 2)\), \((20, 1)\) 2. **For \(d^{2}\) term:** \(nq = 6\) Possible pairs: \((3, 2)\), \((2, 3)\), \((6, 1)\), \((1, 6)\) **Step 2: Find the Correct Combination** Let's try the pair \((5a, 4a)\) for the \(a^{2}\) term. - Let \(m = 5\) and \(p = 4\). Now, we need to find \(n\) and \(q\) such that: \[ 5q + 4n = -23 \quad \text{and} \quad nq = 6 \] Trying \(n = -2\) and \(q = -3\): \[ 5(-3) + 4(-2) = -15 - 8 = -23 \quad \text{and} \quad (-2)(-3) = 6 \] This combination satisfies both conditions. **Step 3: Write the Factored Form** \[ 20a^{2} - 23ad + 6d^{2} = (5a - 2d)(4a - 3d) \] **Answer:** **A.** \[ 20 a^{2}-23 a d+6 d^{2}=(5a-2d)(4a-3d) \]