We are given the equation x3+6x2+5x−42=0. Let's check if x=2 is a root by plugging it in: (2)3+6×(2)2+5×2−42=8+24+10−42=42−42=0 Since the equation equals zero, x=2 is a real root. So, α=2. Now, we substitute α=2 into the matrix: [
α−1
α+1
α+2
α−2
α+3
α−3
α+4
α−4
α+5
]=[
1
3
4
0
5
−1
6
−2
7
] Next, let's find the determinant of this matrix: ‌=|
