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f(x)=2(x^2+2x+1)-8
f(x)=2x^2+4x+2-8
f(x)=2x^2+4x-6
f(x)=x^2+2x-3
delta=b^2-4ac=4+12=16
x1=-2-4/2=-3
x2=-2+4/2=1
Postać iloczynowa: y=a(x-x1)(x-x2)
według wzoru i tego co wyszło to a=1
y=(x+3)(x-1)