Question Factor completely. \[ x^{3}-7 x^{2}-6 x+42 \]

To factor the cubic polynomial \( x^{3}-7 x^{2}-6 x+42 \) completely, we can start by searching for rational roots using the Rational Root Theorem. Testing out values like \( x = 6 \) reveals that \( x = 6 \) is indeed a root! Thus, we can factor out \( (x - 6) \). Long division or synthetic division will show that dividing \( x^{3}-7 x^{2}-6 x+42 \) by \( (x - 6) \) gives us the quadratic \( x^{2}-x-7 \). Now, we need to factor \( x^{2}-x-7 \). This quadratic doesn’t factor neatly over the integers, but we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 28}}{2} = \frac{1 \pm \sqrt{29}}{2}. \] Putting it all together, the complete factorization is: \[ (x - 6)\left(x - \frac{1 + \sqrt{29}}{2}\right)\left(x - \frac{1 - \sqrt{29}}{2}\right). \] So, \( x^{3}-7 x^{2}-6 x+42 \) factors into these three parts!