t² + t + 1 nie da się rozłożyć bo Δ = 1 - 4*1*1 = - 3 < 0 ------------------------------------ b) x⁴ - 5 x³ + 6 x² = x²*(x² -5x +6) = x²*( x -2)*(x -3) bo
Δ = 25 -4*1*6 = 25 - 24 = 1 x = [5 -1]/2 = 2 lub x = [ 5 +1]/2 = 3 ====================================================
(t²-2)(t³-1)
(t-√2)(t+√2)(t³-1)
2) x^4-2x³-3x³+6x²
x³(x-2)-3x²(x-2)
(x-2)(x³-3x²)
(x-2)(x²(x-3))
t⁵ - 2 t³ - t² + 2 = t³*(t²- 2) -(t² -2 ) = (t³ - 1)*(t² - 2) =
= (t - 1)*(t² +t +1)(t-√2)*(t - √2)
t² + t + 1 nie da się rozłożyć bo
Δ = 1 - 4*1*1 = - 3 < 0
------------------------------------
b)
x⁴ - 5 x³ + 6 x² = x²*(x² -5x +6) = x²*( x -2)*(x -3)
bo
Δ = 25 -4*1*6 = 25 - 24 = 1
x = [5 -1]/2 = 2 lub x = [ 5 +1]/2 = 3
====================================================