8. 129
c)
[tex](4x-1)^{2}x+(1-4x)x^{2}-(4x-1)=\\\\=(4x-1)^{2}x - (4x-1)x^{2}-(4x-1)=\\\\=(4x-1)[(4x-1)x - x^{2}-1]=\\\\=(4x-1)(4x^{2}-x-x^{2}-1)=\\\\=(4x-1)(3x^{2}-x-1)[/tex]
Sprawdźmy, czy trójmian 3x² - x - 1 = 0 ma pierwiastki rzeczywiste
[tex]\Delta = (-1)^{2}-4\cdot3\cdot(-1)= 1+12 = 13\\\\\sqrt{\Delta} = \sqrt{13}\\\\x_1 = \frac{1-\sqrt{13}}{2\cdot3} = \frac{1-\sqrt{13}}{6}\\\\x_2 = \frac{1+\sqrt{13}}{2\cdot3} = \frac{1+\sqrt{13}}{6}\\\\zatem\\\\(4x-1)x + (1-4x)x^{2}-(4x-1) = 3(4x-1)(x - \frac{1-\sqrt{13}}{6})(x-\frac{1+\sqrt{13}}{6})[/tex]
e)
[tex](-1-x)(x^{2}-1)-(x+1)^{2}=\\\\=-(1+x)(x-1)(x+1)-(x+1)^{2}=\\\\=-(x+1)^{2}(x-1)-(x+1)^{2}=\\\\=-(x+1)^{2}(x-1 +1)=\\\\=-(x+1)^{2}\cdot x = -x(x+1)^{2}[/tex]
