Odpowiedź :
[tex]zad.4\\\\W(x)= x^3 - x^2 - 9x +9=x^2(x-1)-9(x-1)=(x-1)(x^2-9)=\\\\=(x-1)(x-3)(x+3) \\\\\\pierwiastki \ tego\ trojmianu :\\\\(x-1)(x-3)(x+3)=0\\\\x-1=0\ \ \vee \ \ x-3=0\ \ \vee \ \ x+3=0\\\\x-1=0\ \ \vee \ \ x=3\ \ \vee \ \ x=-3[/tex]


