Question

How many real solutions does the equation $x^7+14x^5+16x^3+30x-560=0$ has?

• A. $3$
• B. $5$
• C. $7$
• D. $1$

Answer 1

The correct option is D $1$

Let $f\left(x\right)=x^7+14x^5+16x^3+30x-560$

$\Rightarrow f^{\prime }\left(x\right)=7x^6+70x^4+48x^2+30>0$ $\forall x\in R$

$\therefore f$ is increasing also $\underset{x\to \infty }{lim}f\left(x\right)=\infty$ ; $\underset{x\to -\infty }{lim}f\left(x\right)=-\infty$

Hence, $f\left(x\right)=0$ has exactly one real root .