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x³+x²
--------------kreska jak w odejmowaniu pisemnym
x²(a+1)+5x
-x²(a+1)-x(a+1)
-------------------
-x(a-4)-b
x(a-4)+a-4
-----------------
-b+a-4 reszta z dzielenia wielomanu
-b+a-4=-8 stąd
a=b-4
x=3pierwiastek równania, zatem -27+9a+15-b=0, 9*(b-4)-12-b=0
9b-b=48
8b=48
b=6
a=6-4=2
mamy wielomian:-x³-2x²+5x-6≤0
(x-3)(-x²-x+2)≤0
-(x-3)(x-1)(x+2)≤0
(x-3)(x-1)(x+2)≥0
X∈(-nieskończoności, -2>∨<1,3>