Answer:D.) 3, √2, −√2
Step-by-step explanation:
Given polynomial: [tex]f(x)=x^3-3x^2-2x+6[/tex]
Let's check all the given options, by substituting given value on x.
A.) −3, −√2, −√2
[tex]f(-3)=(-3)^3-3(-3)^2-2(-3)+6=-27-27+6+6=-42\neq 0[/tex]
⇒ -3 is not a zero of given polynomial.
Thus, this not the required answer.
B.) 3, −√2, −√2
[tex]f(3)=(3)^3-3(3)^2-2(3)+6=27-27-6+6=0[/tex]
[tex]f(-\sqrt{2})=(-\sqrt{2})^3-3(-\sqrt{2})^2-2(-\sqrt{2})+6\\\\=-2\sqrt{2}-6+2\sqrt{2}+6=0[/tex]
But by Descartes rule of sigs , f(x) have 2 positive and 1 negative root.
Thus, this is not the right answer.
C.) −3, √2, √2
-3 is not a zero of given polynomial.
Thus, this not the required answer.
D.) 3, √2, −√2
[tex]f(-3)=(-3)^3-3(-3)^2-2(-3)+6=-27-27+6+6=-42\neq 0[/tex]
[tex]f(\sqrt{2})=(\sqrt{2})^3-3(\sqrt{2})^2-2(\sqrt{2})+6\\\\=2\sqrt{2}-6-2\sqrt{2}+6=0[/tex]
[tex]f(-\sqrt{2})=(-\sqrt{2})^3-3(-\sqrt{2})^2-2(-\sqrt{2})+6\\\\=-2\sqrt{2}-6+2\sqrt{2}+6=0[/tex]
Thus, this is the right option to have zeroes of f(x).
