???? Haaaaaaanka: x³−13x+12=0

Julek: f(x) = x³−13x+12 f(1) = 0 x² + x − 12 x³−13x+12 : (x−1) −(x³ − x²) x² − 13x + 12 −(x² − x) −12x + 12 −(−12x + 12) ==0 f(x) = (x−1)(x² + x − 12) Δ = 1 + 48 = 49 = 7²

	−1+7
x1 =		= 3
	2

	−1−7
x2 =		= −4
	2

f(x) = (x+4)(x−1)(x−3) Odpowiedź : x∊(−4;1;3)

Eta: x³ −13x +12=0 x³−x − 12x +12=0 x(x²−1) −12( x−1)=0 x(x+1)(x−1) −12( x−1)=0 (x−1)(x²+x−12)=0 ......Δ= 49 √Δ= 7 x=3 v x = −4 (x−1)(x−3)(x+4)=0 odp: x= 1 v x= 3 v x= −4