Zad. 5
a)
w(x) = 18x⁵ - 3x⁴ + 6x³ = 3x³(6x² - x + 2)
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6x² - x + 2
a = 6, b = - 1, c = 2
Δ = (- 1)² - 4 · 6 · 2 = 1 - 48 = - 47 < 0, zatem trójmian kwadratowy nie ma postaci iloczynowej
b)
w(x) = 4x⁵ - 4x⁴ -24x³ = 4x³ · (x² - x - 6) = 4x³(x + 2)(x - 3)
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x² - x - 6
a = 1, b = - 1, c = - 6
Δ = (- 1)² - 4 · 1 · (- 6) = 1 + 24 = 25; √Δ = √25 = 5
[tex]x_1 = \frac{-(-1) - 5}{2 \cdot 1} = \frac{1-5}{2} =\frac{-4}{2} = - 2 \\ x_2 = \frac{-(-1) +5}{2 \cdot 1} = \frac{1+5}{2} =\frac{6}{2} =3[/tex]
Zatem: x² - x - 6 = (x + 2)(x - 3)
c)
w(x) = 4x⁵ + 7x⁴ - 2x³ = x³(4x² + 7x - 2) = x³(x + 2)(4x - 1)
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4x² + 7x - 2
a = 4, b = 7, c = - 2
Δ = 7² - 4 · 4 · (- 2) = 49 + 32 = 81; √Δ = √81 = 9
[tex]x_1 = \frac{-7-9}{2 \cdot 4} = \frac{-16}{8} = - 2 \\ x_2 = \frac{-7+9}{2 \cdot 4} = \frac{2}{8} =\frac{1}{4}[/tex]
4x² + 7x - 2 = 4(x + 2)(x - ¹/₄) = (x + 2)(4x - 1)
d)
w(x) = x⁴ - ³/₂ x³ - x² = x²( x² - ³/₂ x - 1) = x²( x² - 1,5x - 1) = x²(x + 0,5)(x - 2)
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x² - 1,5x - 1
a = 1, b = - 1,5, c = - 1
Δ = (1,5)² - 4 · 1 · (- 1) = 2,25 + 4 = 6,25; √Δ = √6,25 = 2,5
[tex]x_1 = \frac{-(-1,5) - 2,5}{2 \cdot 1} =\frac{1,5 - 2,5}{2} = \frac{-1}{2} = - \frac{1}{2} = - 0,5 \\ x_2= \frac{-(-1,5) + 2,5}{2 \cdot 1} =\frac{1,5 + 2,5}{2} = \frac{4}{2} =2[/tex]
x² - 1,5x - 1 = (x + 0,5)(x - 2)
e)
w(x) = - 2x⁷ + x⁶ - x⁵ = - x⁵(2x² - x + 1)
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2x² - x + 1
a = 2, b = - 1, c = 1
Δ = (- 1)² - 4 · 2 · 1 = 1 - 8 = - 7 < 0, zatem trójmian kwadratowy nie ma postaci iloczynowej
f)
w(x) = x⁵ - 2x⁴ + √2x³ = x³(x² - 2x + √2)
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x² - 2x + √2
a = 1, b = - 2, c = √2
Δ = (- 2)² - 4 · 1 · √2 = 4 - 4√2 < 0, zatem trójmian kwadratowy nie ma postaci iloczynowej
