Question

Which of the following represents the zeros of f(x) = x3 − 3x2 − 2x + 6?

Answer 1

D. {3, √ 2, -√ 2}. 
The cubic factors- (x - 3)(x² - 2) = 0.

Answer 2

Answer:D.) 3, √2, −√2

Step-by-step explanation:

Given polynomial: $f(x)=x^3-3x^2-2x+6$

Let's check all the given options, by substituting given value on x.

A.) −3, −√2, −√2

$f(-3)=(-3)^3-3(-3)^2-2(-3)+6=-27-27+6+6=-42\neq 0$

⇒ -3 is not a zero of given polynomial.

Thus, this not the required answer.

B.) 3, −√2, −√2

$f(3)=(3)^3-3(3)^2-2(3)+6=27-27-6+6=0$

$f(-\sqrt{2})=(-\sqrt{2})^3-3(-\sqrt{2})^2-2(-\sqrt{2})+6\\\\=-2\sqrt{2}-6+2\sqrt{2}+6=0$

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

$f(-3)=(-3)^3-3(-3)^2-2(-3)+6=-27-27+6+6=-42\neq 0$

$f(\sqrt{2})=(\sqrt{2})^3-3(\sqrt{2})^2-2(\sqrt{2})+6\\\\=2\sqrt{2}-6-2\sqrt{2}+6=0$

$f(-\sqrt{2})=(-\sqrt{2})^3-3(-\sqrt{2})^2-2(-\sqrt{2})+6\\\\=-2\sqrt{2}-6+2\sqrt{2}+6=0$

Thus, this is the right option to have zeroes of f(x).