\begin{tabular}{l} Factor completely. \\ \( 6 x^{2}-21 x-27 \) \\ Submit \\ \hline\end{tabular}

To factor \( 6x^{2} - 21x - 27 \) completely, we can start by looking for a common factor in all the terms. Notice that 3 is a common factor, so we factor that out first: \[ 3(2x^{2} - 7x - 9) \] Next, we need to factor the quadratic \( 2x^{2} - 7x - 9 \). To do this, we can either use the quadratic formula or look for two numbers that multiply to \( 2 \times -9 = -18 \) and add up to -7. The numbers -9 and 2 fit this, so we can rewrite the middle term: \[ 2x^{2} - 9x + 2x - 9 \] Now, we can group the terms: \[ (2x^{2} - 9x) + (2x - 9) \] Factoring by grouping gives us: \[ x(2x - 9) + 1(2x - 9) \] Thus, we can factor out \( (2x - 9) \): \[ (2x - 9)(x + 1) \] Putting it all together, we factor the original expression as: \[ 3(2x - 9)(x + 1) \] So, the completely factored form of \( 6x^{2} - 21x - 27 \) is: \[ 3(2x - 9)(x + 1) \]