Szczegółowe wyjaśnienie:
[tex]e)\ w(x)=x^3-\sqrt2 x^2+\dfrac{1}{2}x=x\bigg(x^2-\sqrt2 x+\dfrac{1}{2}\bigg)\\\\x^2-\sqrt2 x+\dfrac{1}{2}\\a=1,\ b=-\sqrt2,\ c=\dfrac{1}{2}\\\Delta=(-\sqrt2)^2-4\cdot1\cdot\dfrac{1}{2}=2-2=0\\\\x_0=\dfrac{-(-\sqrt2)}{2\cdot1}=\dfrac{\sqrt2}{2}\\\\x^2-\sqrt2 x+\dfrac{1}{2}=\left(x-\dfrac{\sqrt2}{2}\right)^2[/tex]
[tex]\huge\boxed{w(x)=x\left(x-\dfrac{\sqrt2}{2}\right)^2}[/tex]
[tex]f)\ w(x)=x^5-2x^4-2x^3=x^3(x^2-2x-2)\\\\x^2-2x-2\\a=1,\ b=-2,\ c=-2\\\Delta=(-2)^2-4\cdot1\cdot(-2)=4+8=12\\\sqrt\Delta=\sqrt{12}=\sqrt{4\cdot3}=2\sqrt3\\\\x_1=\dfrac{-(-2)-2\sqrt3}{2\cdot1}=\dfrac{2-2\sqrt3}{2}=1-\sqrt3\\\\x_2=\dfrac{-(-2)+2\sqrt3}{2\cdot1}=\dfrac{2+2\sqrt3}{2}=1+\sqrt3[/tex]
[tex]w(x)=x^3\bigg(x-(1-\sqrt3)\bigg)\bigg(x-(1+\sqrt3)\bigg)\\\\\huge\boxed{w(x)=x^3(x-1+\sqrt3)(x-1-\sqrt3)}[/tex]
