d)
x⁴ - x³ - 7x² + 13x - 6 = 0
W(x) = x⁴ - x³ - 7x² + 13x - 6
D₆ = {1, - 1, 2, - 2, 3, - 3, 6, - 6}
W(1) = 1⁴ - 1³ - 7 · 1² + 13 · 1 - 6 = 1 - 1 - 7 + 13 - 6 = 0
x = 1 jest pierwiastkiem wielomianu W(x)
(x⁴ - x³ - 7x² + 13x - 6) : (x - 1) = x³ - 7x + 6
- x⁴ + x³
- 7x² + 13x - 6
7x² - 7x
6x - 6
- 6x + 6
0
Zatem: W(x) = x⁴ - x³ - 7x² + 13x - 6 = (x - 1)(x³ - 7x + 6)
U(x) = x³ - 7x + 6
D₆ = {1, - 1, 2, - 2, 3, - 3, 6, - 6}
U(1) = 1³ - 7 · 1 + 6 = 1 - 7 + 6 = 0
x = 1 jest pierwiastkiem wielomianu U(x)
(x³ - 7x + 6) : (x - 1) = x² + x - 6
- x³ + x²
x² - 7x + 6
- x² + x
- 6x + 6
6x - 6
0
Zatem: U(x) = x³ - 7x + 6 = (x - 1)(x² + x - 6)
oraz W(x) = x⁴ - x³ - 7x² + 13x - 6 = (x - 1)(x³ - 7x + 6) = (x - 1)(x - 1)(x² + x - 6) =
= (x - 1)²(x² + x - 6)
Stąd:
x⁴ - x³ - 7x² + 13x - 6 = 0
(x - 1)²(x² + x - 6) = 0
(x - 1)² = 0 lub x² + x - 6 = 0
(x - 1)² = 0
x - 1 = 0
x = 1
x² + x - 6 = 0
Δ = 1² - 4 · 1 · (- 6) = 1 + 24 = 25; √Δ = √25 = 5
[tex]x_1 = \frac{-1-5}{2 \cdot 1} =\frac{-6}{2}=-3 \\ x_2 = \frac{-1+5}{2 \cdot 1} =\frac{4}{2}=2 \\ \underline{x = - 3 \ lub \ x = 2}[/tex]
Odp. x = - 3 lub x = 1 lub x = 2
e)
x⁴ - 4x³ + 6x² - 4x + 1 = 0
W(x) = x⁴ - 4x³ + 6x² - 4x + 1
D₁ = {1, - 1}
W(1) = 1⁴ - 4 · 1³ + 6 · 1² - 4 · 1 + 1 = 1 - 4 + 6 - 4 + 1 = 0
x = 1 jest pierwiastkiem wielomianu W(x)
(x⁴ - 4x³ + 6x² - 4x + 1) : (x - 1) = x³ - 3x² + 3x - 1
- x⁴ + x³
- 3x³ + 6x² - 4x + 1
3x³ - 3x²
3x² - 4x + 1
- 3x² + 3x
- x + 1
x - 1
0
Zatem: W(x ) = x⁴ - 4x³ + 6x² - 4x + 1 = (x - 1)(x³ - 3x² + 3x - 1) = (x - 1)(x - 1)³ =
= (x - 1)⁴
Stąd:
x⁴ - 4x³ + 6x² - 4x + 1 = 0
(x - 1)⁴ = 0
x - 1 = 0
x = 1
Odp. x = 1
f)
x⁴ + 3x³ - 5x² - 12x + 4 = 0
W(x) = x⁴ + 3x³ - 5x² - 12x + 4
D₄ = {1, - 1, 2, - 2, 4, - 4}
W(1) = 1⁴ + 3 · 1³ - 5 · 1² - 12 · 1 + 4 = 1 + 3 - 5 - 12 + 4 = - 9 ≠ 0
W(- 1) = (- 1)⁴ + 3 · (- 1)³ - 5 · (- 1)² - 12 · (- 1) + 4 = 1 - 3 - 5 + 12 + 4 = 9 ≠ 0
W(2) = 2⁴ + 3 · 2³ - 5 · 2² - 12 · 2 + 4 = 16 + 24 - 20 - 24 + 4 = 0
x = 2 jest pierwiastkiem wielomianu W(x)
(x⁴ + 3x³ - 5x² - 12x + 4) : (x - 2) = x³ + 5x² + 5x - 2
- x⁴ + 2x³
5x³ - 5x² - 12x + 4
- 5x³ + 10x²
5x² - 12x + 4
- 5x² + 10x
- 2x + 4
2x - 4
0
Zatem: W(x ) = x⁴ + 3x³ - 5x² - 12x + 4 = (x - 2)(x³ + 5x² + 5x - 2)
U(x) = x³ + 5x² + 5x - 2
D₋₂ = {1, - 1, 2, - 2}
U(1) = 1³ + 5 · 1² + 5 · 1 - 2 = 1 + 5 + 5 - 2 = 9 ≠ 0
U(- 1) = (- 1)³ + 5 · (- 1)² + 5 · (- 1) - 2 = - 1 + 5 - 5 - 2 = - 3 ≠ 0
U(2) = 2³ + 5 · 2² + 5 · 2 - 2 = 8 + 20 + 10 - 2 = 36 ≠ 0
U(- 2) = = (- 2)³ + 5 · (- 2)² + 5 · (- 2) - 2 = - 8 + 20 - 10 - 2 = 0
x = - 2 jest pierwiastkiem wielomianu U(x)
(x³ + 5x² + 5x - 2) : (x + 2) = x² + 3x - 1
- x³ - 2x²
3x² + 5x - 2
- 3x² - 6x
- x - 2
x + 2
0
Zatem: U(x) = x³ + 5x² + 5x - 2 = (x + 2)(x² + 3x - 1)
oraz W(x) = x⁴ + 3x³ - 5x² - 12x + 4 = (x - 2)(x + 2)(x² + 3x - 1)
Stąd:
x⁴ + 3x³ - 5x² - 12x + 4 = 0
(x - 2)(x + 2)(x² + 3x - 1) = 0
x - 2 = 0 lub x + 2 = 0 lub x² + 3x - 1 = 0
x - 2 = 0
x = 2
x + 2 = 0
x = - 2
x² + 3x - 1 = 0
Δ = 3² - 4 · 1 · (- 1) = 9 + 4 = 13; √Δ = √13
[tex]x_1=\frac{-3-\sqrt{13}}{2 \cdot 1} =\frac{-3-\sqrt{13}}{2} \\ x_2=\frac{-3+\sqrt{13}}{2 \cdot 1} =\frac{-3+\sqrt{13}}{2} \\ \underline{x =\frac{-3-\sqrt{13}}{2} \ lub \ x = \frac{-3+\sqrt{13}}{2}}[/tex]
Odp. [tex]x =\frac{-3-\sqrt{13}}{2} \ lub \ x = - 2 \ lub \ x = \frac{-3+\sqrt{13}}{2} \ lub \ x = 2[/tex]
