Question

For the expression $\frac{4x^6+7x^4+5x^2}{x^n}$ to be polynomial, what could be the possible value(s) of '$n$' $(n\ge 0)?$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A), (B), (C)

$2$

Let's simplify the given expression,
$\frac{4x^6+7x^4+5x^2}{x^n}$

$=4x^{(6-n)}+7x^{(4-n)}+5x^{(2-n)}$ $[\because \frac{a^m}{a^n}=a^{(m-n)}]$

As we all know, for an expression to be polynomial, the exponents of variables must be whole numbers.

For the expression, $4x^{(6-n)}+7x^{(4-n)}+5x^{(2-n)}$ to be polynomial, the exponents $(6-n),(4-n)$ and $(2-n)$ must belong to set of whole numbers.

For $(6-n),(4-n)$ and $(2-n)$ to be whole numbers, $n$ must be an integer.

$(6-n)$ and $(4-n)$ must be whole numbers, if $(2-n)$ is a whole number.

Therefore, for the expression, $4x^{(6-n)}+7x^{(4-n)}+5x^{(2-n)}$ to be polynomial, $2-n\ge 0\Rightarrow n-2\le 0\Rightarrow n\le 2$
$\therefore$ Positive integers less than or equals to $2$ are $\{0,1,2\}.$

As a result, possible values of $\underset{–}{n}$ could be $\{0,1,2\}.$