Answer:
There are 3 zeros, which are:
[tex]x=7,\:x=-\sqrt{2},\:x=\sqrt{2}[/tex]
Step-by-step explanation:
Given the function
[tex]f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14[/tex]
To get the zeros of [tex]f(x)[/tex], set [tex]y[/tex] or [tex]f(x) =0[/tex]
so
[tex]0=x^3-7x^2-2x+14[/tex]
as
[tex]x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)[/tex]
so
[tex]\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0[/tex]
Using the zero factor principle:
[tex]\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
so
[tex]x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0[/tex]
solving
[tex]x-7=0[/tex]
[tex]x+\sqrt{2}=0[/tex]
[tex]x-\sqrt{2}=0[/tex]
Therefore, there are 3 zeros, which are:
[tex]x=7,\:x=-\sqrt{2},\:x=\sqrt{2}[/tex]
