Szczegółowe wyjaśnienie:
[tex]6x^4+3x^3-3x^2=3x^2(2x^2+x-1)=3x^2(2x^2+2x-x-1)\\\\=3x^2\bigg[2x(x+1)-1(x+1)\bigg]=3x^2(x+1)(2x-1)[/tex]
Można ten trójmian kwadratowy w nawiasie rozłożyć za pomocą Delty.
[tex]4x^6-12x^5+x^4=x^4(4x^2-12x+1)\qquad(*)\\\\4x^2-12x+1=0\\a=4;\ b=-12;\ c=1\\\\\Delta=b^2-4ac\\\Delta=(-12)^2-4\cdot4\cdot1=144-16=128\\\sqrt\Delta=\sqrt{128}=\sqrt{64\cdot2}=8\sqrt2\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-(-12)-8\sqrt2}{2\cdot4}=\dfrac{12-8\sqrt2}{8}=\dfrac{3-2\sqrt3}{2}\\\\x_2=\dfrac{-(-12)+8\sqrt2}{2\cdot4}=\dfrac{12+8\sqrt2}{8}=\dfrac{3+2\sqrt2}{2}[/tex]
[tex](*)=3x^2\left(x-\dfrac{3-2\sqrt2}{2}\right)\left(x-\dfrac{3+2\sqrt2}{2}\right)[/tex]