Odpowiedź i szczegółowe wyjaśnienie:
[tex]w(x)=-2x^3+20x^2-10x\\\\w(x)=-2x (x^2-10x+5)\\\\[/tex]
-2x=0 = x=0
[tex]x^2-10x+5=0\\\\a=1;\ b=-10;\ c=5\\\\\Delta=b^2-4ac=(-10)^2-4\cdot1\cdot5=100-20=80\\\sqrt\Delta=\sqrt{80}=\sqrt{16\cdot 5}=4\sqrt5\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}=\dfrac{-(-10)-4\sqrt5}{2\cdot1}=\dfrac{10-4\sqrt5}{2}=\dfrac{2(5-2\sqrt5)}{2}=5-2\sqrt5\\\\x_2=\dfrac{-b+\sqrt\Delta}{2a}=\dfrac{-(-10)+4\sqrt5}{2\cdot1}=\dfrac{10+4\sqrt5}{2}=\dfrac{2(5+2\sqrt5)}{2}=5+2\sqrt5[/tex]
Ostatecznie nasz wielomian po rozłożeniu na czynniki ma postać:
[tex]w(x)=-2x(x-(5-2\sqrt5))(x+(5+2\sqrt5)\\\\w(x)=-2x(x-5+2\sqrt5)(x+5+2\sqrt5)[/tex]
[tex]y = (x + 1) (x - 3)\\\\y=x^2-3x+x-3=x^2-2x-3[/tex]
