Question

The additive inverse of $\frac{x^7-7x^2+18}{x^3-2}$ is

• A. $\frac{x^5+7x^2+18}{x^3-2}$
• B. $\frac{-x^5-7x^2+18}{x^3-2}$
• C. $\frac{-x^7+7x^2-18}{x^3-2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{-x^7+7x^2-18}{x^3-2}$

Given that, $\Rightarrow \frac{x^7-7x+18}{x^3-2}$

Now,

$\Rightarrow \frac{x^7-7x+18}{x^3-2}+\frac{-(x^7-7x+18)}{x^3-2}$

$\Rightarrow \frac{x^7-7x+18}{x^3-2}-\frac{(x^7-7x+18)}{x^3-2}=0$

Hence, $\Rightarrow \frac{-(x^7-7x+18)}{x^3-2}=\frac{-x^7+7x-18}{x^3-2}$ is the additive inverse of $\frac{(x^7-7x+18)}{x^3-2}$

Hence, this is the answer.