Persamaan x³-5x²+10x-6=0 memiliki akar-akar persamaan a,b, dan c. nilai dari (a+2)(b+2)(c+2) adalah​

Persamaan Kubik

ax³ + bx² + cx + d = 0

x1 + x2 + x3 = -b/a

x1x2 + x1x3 + x2x3 = c/a

x1x2x3 = -d/a

x³ – 5x² + 10x – 6 = 0

a = 1 ; b = -5 ; c = 10 ; d = -6

x1 + x2 + x3 = -(-5)/1 = 5

x1x2 + x1x3 + x2x3 = 10/1 = 10

x1x2x3 = -(-6)/1 = 6

pada soal akar2 → a, b, c

sy pakai α, в, ς

x1 = α

x2 = в

x2 = ς

(α + 2)(в + 2)(ς + 2)

= αвς + 2(αв + ας + вς) + 4(α + в + ς) + 8

= 6 + 2(10) + 4(5) + 8

= 6 + 20 + 20 + 8

= 54

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cara lain

x³ – 5x² + 10x – 6 = 0

jumlah koefisien dan konstanta = 0

x – 1 adalah salah satu faktor persamaan

1 – 5 + 10 – 6 = 0 ✔

x³ – 5x² + 10x – 6 = 0

(x – 1)(x² – 4x + 6) = 0

x1 = a = 1

x² – 4x + 6 = 0

dg rumus abc :

x2 = b = 2 + √(–2)

x3 = c = 2 – √(–2)

(a + 2)(b + 2)(c + 2)

= (1 + 2)(4 + √2)(4 – √2)

= 3 (4² – (√(-2))²)

= 3 (16 + 2)

= 54