Odpowiedź :
(x-1)(x-1)(x-3)(x²-2x-6)=(x-1)²(x-3)(x²-2x-6)
Δ=-2²x((-6)x(-4))=28
√Δ=√28=2√7
x₁=(2-2√7)/2=1-√7
x₂=(2+2√7)/2=1+√7
(x-1)(x-1)(x-3)(x-1-√7)(x-1+√7)
:) Wszystko schematem Hornera
Δ=-2²x((-6)x(-4))=28
√Δ=√28=2√7
x₁=(2-2√7)/2=1-√7
x₂=(2+2√7)/2=1+√7
(x-1)(x-1)(x-3)(x-1-√7)(x-1+√7)
:) Wszystko schematem Hornera
Znajdź pierwiastki wielomianu
x⁵-7x⁴+11x³+13x²-36x+18<0
W(1)=0
zatem z Hornera
(x-1)(x-1)(x-3)(x²-2x-6)=(x-1)²(x-3)(x²-2x-6)=
=(x-1)(x-1)(x-3)(x-1-√7)(x-1+√7)
x²-2x-6=0
Δ=-2²x[(-6)*(-4)]=28
√Δ=√28=2√7
x₁=(2-2√7)/2=1-√7
x₂=(2+2√7)/2=1+√7
x⁵-7x⁴+11x³+13x²-36x+18<0
W(1)=0
zatem z Hornera
(x-1)(x-1)(x-3)(x²-2x-6)=(x-1)²(x-3)(x²-2x-6)=
=(x-1)(x-1)(x-3)(x-1-√7)(x-1+√7)
x²-2x-6=0
Δ=-2²x[(-6)*(-4)]=28
√Δ=√28=2√7
x₁=(2-2√7)/2=1-√7
x₂=(2+2√7)/2=1+√7