Zad. 1
a)
w(x) = 8x³ + 27 = (2x)³ + 3³ = (2x + 3)(4x² - 6x + 9)
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4x² - 6x + 9
a = 4, b = - 6, c = 9
Δ = (- 6)² - 4 · 4 · 9 = 36 - 144 = - 108 < 0
Trójmian kwadratowy 4x² - 6x + 9 nie ma postaci iloczynowej
b)
w(x) = x³ - 8x² + 15x = x · (x² - 8x + 15) = x(x - 3)(x - 5)
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x² - 8x + 15
a = 1, b = - 8, c = 15
Δ = (- 8)² - 4 · 1 · 15 = 64 - 60 = 4; √Δ = √4 = 2
[tex]x_1 =\frac{-(-8) - 2}{2 \cdot 1} =\frac{8-2}{2} =\frac{6}{2} = 3 \\ x_2 =\frac{-(-8)+ 2}{2 \cdot 1} =\frac{8+2}{2} =\frac{10}{2} = 5[/tex]
x² - 8x + 15 = (x - 3)(x - 5)
c)
w(x) = (x² - 2)² - 36 = (x² - 2)² - 6² = (x² - 2 - 6)(x² - 2 + 6) = (x² - 8)(x² + 4) =
= (x² - (√8)²)(x² + 4) = (x - √8)(x + √8)(x² + 4) = (x - √(4 · 2))(x + √(4 · 2))(x² + 4)=
= (x - 2√2)(x + 2√2)(x² + 4)
