Odpowiedź :
[tex]\frac{10x^3+6(x^2-2)}{2} -2x(3-2x-x^2) = \frac{10x^3+6x^2-12}{2} -6x+4x^2+2x^3 = \\\\ = \frac{\not{2}^1(5x^3+3x^2-6)}{\not{2}_1} -6x+4x^2+2x^3 =5x^3+3x^2-6-6x+4x^2+2x^3 = 7x^3+7x^2-6x-6[/tex]
[tex]\frac{10x^3+6(x^2-2)}{2} -2x(3-2x-x^2) = \frac{10x^3+6x^2-12}{2} -6x+4x^2+2x^3 = \\\\ = \frac{\not{2}^1(5x^3+3x^2-6)}{\not{2}_1} -6x+4x^2+2x^3 =5x^3+3x^2-6-6x+4x^2+2x^3 = 7x^3+7x^2-6x-6[/tex]