Oblicz pierwiastki wielomianu Bezouta
p(x)= x^3+2x^2-5x-6
f(2)=8+8-10-6=0
dzielenie schematem hornera:
1 2 -5 -6
1 4 3 0
x^3+2x^2-5x-6=(x^2+4x+3)(x-2)
p(x)=(x^2+4x+3)(x-2)
Δ=16-12=4
√Δ=2
x1=-4 + 2 / 2 = -2/2= -1
x2=-4 - 2 / 2 = -6/2= -3
oraz z drugiego nawiasu ( x-2 ) wyjdzie x=2
więc:
x=-1 v x =-3 v x=2
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f(2)=8+8-10-6=0
dzielenie schematem hornera:
1 2 -5 -6
1 4 3 0
x^3+2x^2-5x-6=(x^2+4x+3)(x-2)
p(x)=(x^2+4x+3)(x-2)
Δ=16-12=4
√Δ=2
x1=-4 + 2 / 2 = -2/2= -1
x2=-4 - 2 / 2 = -6/2= -3
oraz z drugiego nawiasu ( x-2 ) wyjdzie x=2
więc:
x=-1 v x =-3 v x=2