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f(x)=x³+3x²-4
f(1)=1+3-4=0
schemat Hornera
1 3 0 -4
1 1 4 4 0
f(x)=(x-1)(x²+4x+4)=(x-1)(x-2)²
(x-1)(x-2)²=0
x=1 lub x=2
x³+3x²-4=0
dzielniki 4
+-1 +-2 +-4
a_3 a_2 a_1 a_0
1 3 0 -4
1 1 4 4 0
x²+4x+4=0
Δ=16-16=0 1 rozw
x=-4/2=-2
rozwiazaniem sa -2 i 1
mozna prosi naj ??:>