Respuesta :

[tex]\bf \begin{cases} x=\sqrt{3}\implies &x-\sqrt{3}=0\\ x=-4\implies &x+4=0\\ x=4\implies &x-4=0 \end{cases} \\\\\\ (x-\sqrt{3})(x+4)(x-4)=\stackrel{y}{0}\implies (x-\sqrt{3})(\stackrel{\stackrel{difference}{of~squares}}{x^2-4^2})=y \\\\\\ (x-\sqrt{3})(x^2-16)=y\implies x^3-16x-x^2\sqrt{3}+16\sqrt{3}=y \\\\\\ x^3-x^2\sqrt{3}-16x+16\sqrt{3}=y[/tex]
Kyleesimmons48

Answer:

c

Step-by-step explanation: