Answer:
The zeros for the function
[tex]x=6,\:x=4i,\:x=-4i[/tex]
Step-by-step explanation:
Given the expression
[tex]y=x^3-6x^2+16x-96[/tex]
Plug in y = 0 to determine all the zeros
[tex]x^3-6x^2+16x-96=0[/tex]
as
[tex]x^3-6x^2+16x-96=\left(x-6\right)\left(x^2+16\right)[/tex]
so the expression becomes
[tex]\left(x-6\right)\left(x^2+16\right)=0[/tex]
Using the zero factor principle
if ab=0, then a=0 or b=0 (or both a=0 and b=0)
[tex]x-6=0\quad \mathrm{or}\quad \:x^2+16=0[/tex]
solving
[tex]x-6=0[/tex]
[tex]x = 6[/tex]
solving
[tex]x^2+16=0[/tex]
subtract 16 from both sides
[tex]x^2+16-16=0-16[/tex]
[tex]x^2=-16[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{-16},\:x=-\sqrt{-16}[/tex]
as
[tex]x=\sqrt{-16}[/tex]
[tex]\sqrt{-16}=\sqrt{-1}\sqrt{16}[/tex]
as
[tex]\sqrt{-1}=i[/tex]
so
[tex]x=4i[/tex]
and
[tex]x=-\sqrt{-16}[/tex]
[tex]-\sqrt{-16}=-\sqrt{-1}\sqrt{16}[/tex]
as
[tex]\sqrt{-1}=i[/tex]
so
[tex]x=-4i[/tex]
Thus, the zeros for the function
[tex]x=6,\:x=4i,\:x=-4i[/tex]
