Tentukan akar-akar persamaan polinomial x³-3x²-6x+8=0

tentukan akar-akar persamaan polinomial x³-3x²-6x+8=0

Jawaban yang benar adalah x = -2, x = 1, dan x = 4

Ingat!
Diberikan polinomial p(x) = 0
Jika p(a) = 0, maka x = a merupakan salah satu akar dari polinomial p(x) = 0 dan (x-a) adalah salah satu faktor dari p(x)

Substitusi x = 1 ke x³-3x²-6x+8
1³-3•1²-6•1+8
= 1 – 3 – 6 + 8
= 0
Sehingga x = 1 merupakan akar dari x³-3x²-6x+8 = 0

Perhatikan
x³-3x²-6x+8 = 0
(x-1) (x²-2x-8) = 0
(x-1) (x-4) (x+2) = 0
x-1 = 0 atau x-4 = 0 atau x+2 = 0
x = 1 atau x = 4 atau x = -2

Jadi, akar dari x³-3x²-6x+8 = 0 adalah x = -2, x = 1, dan x = 4

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