salah satu akar dari persamaan [tex] {x}^{3} + {px}^{2} + (2p - 1)x + (p - 8) = 0[/tex] adalah -2. hasil kali ketiga akar akarnya adalah..... tolong di jawab c

[tex]{x}^{3} + {px}^{2} + (2p - 1)x + (p - 8) = 0 \\ \\ f( - 2) = 0 \\ { - 2}^{3} + p { (- 2)}^{2} + (2p - 1)( - 2) + p - 8 = 0 \\ - 8 + 4p - 4p + 2 + p - 8 = 0 \\ p - 14 = 0 \\ p = 14[/tex]
masukan nilai p..
[tex]{x}^{3} + {14x}^{2} + (2(14) - 1)x + (14 - 8) = 0 \\ {x}^{3} + 14 {x}^{2} + 27x + 6 = 0[/tex]
[tex]x = - 2 \\ (x + 2) \\ \\ {x}^{3} + 14 {x}^{2} + 27x + 6 = 0 \\ (x + 2)( {x}^{2} + 12x + 3) = 0 [/tex]
[tex]( {x}^{2} + 12x + 3) = 0 \\ \\ x1.x2 = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} - 4ac} }{2a} \\ = \frac{ - 12 \frac{ + }{} \sqrt{ {12}^{2} - 4 \times 1 \times 3} }{2 \times 1} \\ = \frac{ - 12 \frac{ + }{} \sqrt{132} }{2} \\ = \frac{ - 12 \frac{ + }{}2 \sqrt{33} }{2} = - 6 \frac{ + }{} \sqrt{33} \\ \\ x1 = \sqrt{33} - 6 \\ x2 = - 6 - \sqrt{33} [/tex]
maka:
[tex]x1 \times x2 \times x = ( \sqrt{33} - 6) \times ( - 6 - \sqrt{33} ) \times ( - 2) \\ = ( \sqrt{33} - 6) \times (12 + 2 \sqrt{33} ) \\ = 12 \sqrt{33} + 66 - 72 - 12 \sqrt{33} \\ = ( - 6)[/tex]


jadi hasil kali ketiga akarnya adalah (-6)...