Question

Select the correct statements for the equation $x^5+4x^4-3x^2+x-6=0.$

• A. The number of negative roots can be $0$
• B. The number of negative roots can be $2$
• C. The number of negative roots can be $1$
• D. The number of negative roots can be $3$

Answer 1

Answer: (A), (B)

The number of negative roots can be $2$

Let $f\left(x\right)=x^5+4x^4-3x^2+x-6$.
$\Rightarrow f(-x)=(-x{)}^5+4(-x{)}^4-3(-x{)}^2+(-x)-6$
$\Rightarrow f(-x)=-x^5+4x^4-3x^2-x-6$
The number of sign changes in $f(-x)$ is $2$.
$\Rightarrow$ The maximum possible no. of negative real roots is $2.$
$\Rightarrow$ The possible number of negative real roots can be $0$ or $2$.