Respuesta :

WojtekR
[tex]f(x)=x^3-x^2+6x-6=x^2(x-1)+6(x-1)=(x^2+6)(x-1)=\\\\=\big(x^2-(-6)\big)(x-1)= \big(x^2-(-1\cdot6)\big)(x-1)=\\\\=\big(x^2-i^2(\sqrt{6})^2\big)(x-1)= \big(x^2-(i\sqrt{6})^2\big)(x-1)=\\\\=\boxed{(x-i\sqrt{6})(x+i\sqrt{6})(x-1)}[/tex]

So complex zeros:

[tex]x_1=-i\sqrt{6}[/tex]

and

[tex]x_2=i\sqrt{6}[/tex]

As options are not given, you can match your answer with the option.

The equation of the polynomial is , [tex]f(x)=x^{3}-x^{2}+6 x -6[/tex]

       =  x²(x-1)+6(x-1)

       = (x² +6)(x-1)

       = (x+ i√6)(x-i√6)(x-1)→a²+b²=(a+ib)(a-ib), i²=-1

The two complex zeroes are , put x+i√6=0 ∧ x-i√6=0 is -i√6 and i√6.

Ver imagen Аноним

Otras preguntas