Question

The minimum number of zeroes with real values of the polynomial $a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

$a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex$ can be written as $x(a{x}^4+b{x}^3+c{x}^2+dx+e)$

For finding roots of the given equation $:$

$x(a{x}^4+b{x}^3+c{x}^2+dx+e)=0$

$\Rightarrow k\left(x\right)(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )=0$

Atleast one real root $x=0$ exists.

We can't comment on other roots with given data

$\therefore$ Minimum number of zeroes with real values is $1$