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1 -1 -5 7 10
-1 0 -1 2 3 -10
1 -2 -3 10 =
(z-1)(z³-2z²-3z+10)=0
z=1 ∨ z³-2z²-3z+10=0⇔(z+2)(z²-4z+5)=0⇔z=-2 ∨ z²-4z+5=0
1 -2 -3 10 Δ=-4=4*i² , √Δ=2i
-2 0 -2 8 -10 z=4-2i)/2 ∨ z=(4+2i)/2
1 -4 5 = z=2-i ∨ z=2+i
b)
1 3 1 13 30
-2 0 -2 -2 2 -30
1 1 -1 15 =
(z+2)(z³+z²-z+15)=0
z=-2 ∨ z³+z²-z+15=0⇔(z+3)(z²-2z+5)=0⇔z=-3 ∨ z²-2z+5=0
Δ=-16=16*i², √Δ=4i
1 1 -1 15 z=(2-4i)/2 ∨ z=(2+4i)/2
-3 0 -3 6 -15 z=1-2i ∨ z=1+2i
1 -2 5 =