Hello!
The answer is: C) [tex]x^{3}+2x^{2}-3x-6[/tex]
Why?
Factoring we have:
[tex]x^{3}+2x^{2}-3x-6=(x^{2})(x+2)-3(x+2)\\(x^{2})(x+2)-3(x+2)=(x+2)(x^{2}-3)\\(x+2)(x+\sqrt{3})(x-\sqrt{3})=0[/tex]
So,
[tex]x_{1}=-2\\x_{2}=-\sqrt{3}\\x_{3}=\sqrt{3}[/tex]
Let's substitute the roots into the equation,
Substituting: -2
[tex](-2+2)(x+\sqrt{3})(x-\sqrt{3})=(0)(x+\sqrt{3})(x-\sqrt{3})=0[/tex]
Substituting: [tex]-\sqrt{3}[/tex]
[tex](x+2)*(-\sqrt{3}+\sqrt{3})*(x-\sqrt{3})=(x+2)*(0)*(x-\sqrt{3})=0[/tex]
Substituting: [tex]\sqrt{3}[/tex]
[tex](x+2)*(x+\sqrt{3})*(\sqrt{3}-\sqrt{3})=(x+2)*(x+\sqrt{3})*(0)=0[/tex]
Have a nice day!
