Question

The mximum number of real roots of the equation $f\left(x\right)=x^6+8x^2-14x+1$ is

• A. $0$
• B. $1$
• C. $6$
• D. $2$

Answer 1

The correct option is D $2$

Given polynomial is $f\left(x\right)=x^6+8x^2-14x+1$

According to Descartes’ Rule of Signs the number of positive real zeros is either equal to the number of sign changes of $f\left(x\right)$ or is less than the number of sign changes by an even integer.

In the given $f\left(x\right)$ we have two sign changes, so there are either two or zero positive real roots.

According to Descartes’ Rule of Signs the number of negative real zeros is either equal to the number of sign changes of $f(-x)$ or is less than the number of sign changes by an even integer.

$f(-x)=x^6+8x^2+14x+1$
In the given $f(-x)$ we have zero sign change, so there can be no negative real root.
$\Rightarrow$ total maximum possible number of real roots $=2+0=2$