Question

Which polynomial will definitely have nonreal zeros?

Answer 1

Answer:

Option C - $f(x) =x^3-2x^2$  

Step-by-step explanation:

To find : Which polynomial will definitely have non-real zeros?

Solution :

Descartes' rule of sign is used to determine the number of real zeros of a polynomial function.

Now, we apply Descartes's  rule in each polynomial.

A) $f(x) =x^3-2x^2-x-2$

(+ - - - ) f(x) has one sign change,

So, One positive roots.

$f(-x) =(-x)^3 - 2(-x)^2 -(-x) - 2$

$f(-x) = -x^3-2x^2+x-2$

(- - + -) f(-x) has two sign change,

So, there are either 0 or 2 negative real roots.

B) $f(x) =x^3-2x^2+x-2$

(+ - + - ) f(x) has three sign change,

So, Three positive roots.

$f(-x) =(-x)^3 - 2(-x)^2 +(-x) - 2$

$f(-x) = -x^3-2x^2-x-2$

(- - - -) f(-x) has no sign change,

So, there are no negative real roots.

C) $f(x) =x^3-2x^2$

(+ -) f(x) has one sign change,

So, One positive roots.

$f(-x) =(-x)^3 - 2(-x)^2$

$f(-x) = -x^3-2x^2$

(- -) f(-x) has no sign change,

So, there are no negative real roots.

Since, the order of polynomial is three and we get only one real root so the roots left possibly be imaginary or non-real zeros.

D) $f(x) =x^4-8x^2+16$

(+ - +) f(x) has two sign change,

So, two positive roots.

$f(-x) =(-x)^4 - 8(-x)^2 +16$

$f(-x) = x^4-8x^2-x+16$

(+ - +) f(-x) has two sign change,

So, there are two negative real roots.

Therefore, Only Option C will have non-real zeros.

Answer 2

The best and most correct answer among the choices provided from your question about nonreal zeroes is the fourth option or letter D.X^4-8x^2+16. It was found out that the zeroes of this polynomial have one positive real root which is nonzero because it has sign change. I hope that has come to your help.