Jawab:
1. x = ½ (1 ± i √11)
2. (3x - 1) dan (x + 1)
Penjelasan dengan langkah-langkah:
Semua diselesaikan dengan metode Horner
Nomor 1
x³ - 3x² + 5x - 6 = 0
(x - 2)(x² - x + 3) = 0
• x - 2 = 0 → x = 2, x ∈ R
• x² - x + 3 = 0
[tex]\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(3)}}{2(1)}\\&=\frac{1\pm\sqrt{-11}}{2}\\&=\frac{1\pm\sqrt{-1}\sqrt{11}}{2}\\&=\frac{1\pm i\sqrt{11}}{2}\end{aligned}[/tex]
Nomor 2
3x⁴ - 4x³ + x² + 6x - 2
= (x - ⅓)(x + 1)(3x² - 6x + 6)
= (x - ⅓)(x + 1) 3(x² - 2x + 2)
= (3x - 1)(x + 1)(x² - 2x + 2)
x² - 2x + 2 merupakan faktor kuadrat
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Verified answer
Jawab:
1. x = ½ (1 ± i √11)
2. (3x - 1) dan (x + 1)
Penjelasan dengan langkah-langkah:
Semua diselesaikan dengan metode Horner
Nomor 1
x³ - 3x² + 5x - 6 = 0
(x - 2)(x² - x + 3) = 0
• x - 2 = 0 → x = 2, x ∈ R
• x² - x + 3 = 0
[tex]\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(3)}}{2(1)}\\&=\frac{1\pm\sqrt{-11}}{2}\\&=\frac{1\pm\sqrt{-1}\sqrt{11}}{2}\\&=\frac{1\pm i\sqrt{11}}{2}\end{aligned}[/tex]
Nomor 2
3x⁴ - 4x³ + x² + 6x - 2
= (x - ⅓)(x + 1)(3x² - 6x + 6)
= (x - ⅓)(x + 1) 3(x² - 2x + 2)
= (3x - 1)(x + 1)(x² - 2x + 2)
x² - 2x + 2 merupakan faktor kuadrat