Q:

Which answer best describes the complex zeros of the polynomial function? f(x)=x3βˆ’x2+6xβˆ’6?

Accepted Solution

A:
[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]