Respuesta :
[tex]f(x)=(x-\sqrt6)(x+\sqrt6)(x+3)=(x^2-6)(x+3)\\\\=\boxed{x^3+3x^2-6x-18}[/tex]
Answer:
[tex]x^{3} +3x^{2} -6x-18[/tex]
Step-by-step explanation:
We are given zeroes and we need to find a cubic function with these zeroes
zeroes given are : [tex]\sqrt{6} , - \sqrt{6} , -3[/tex]
Since these are zeroes so these can be written as:
[tex](x-\sqrt{6} )(x-(-\sqrt{6} ))(x-(-3)) = 0[/tex]
⇒[tex](x-\sqrt{6} )(x+\sqrt{6} )(x+3) = 0[/tex]
⇒[tex](x^{2} +\sqrt{6} x- \sqrt{6} x -6)(x+3)[/tex]
⇒[tex](x^{2} -6)(x+3)[/tex]
⇒[tex]x^{3} +3x^{2} -6x-18[/tex]
Thus the required cubic function is [tex]x^{3} +3x^{2} -6x-18[/tex]