The number of real roots of $2x^5+8x-7$ Is there a (theoretical) method to tell the number of real roots of $$2x^5+8x-7=0?$$
Note, without using calculator.
 A: The polynomial is increasing. It is negative at $x=0$ and positive at $x=1$, thus it has exactly one real root, and it's somewhere between those two points.
A: Using Descartes' Sign Rule,
if $f(x)=2x^5+8x-7$
Since there is only one sign change, there is a maximum of one possible positive root,so has exactly one positive root as the degree of $f(x)$ is $3$ with real coefficients,so complex roots will occur in pair.
For $f(-x)=-2x^5-8x-7$
Since there is no sign change, there is no possible negative root for $f(x)$.
So, the number of real roots is $1$.
A: Yes, this is a pretty standard MVT+ Rolle Theorem problem...
Hint: Let $f(x)=2x^5+8x-7$. Then $f'(x)=10x^4+8$. How many roots does $f'$ have? Ho many roots could then have $f$?
Can you use the IVT to finish the problem?
A: We can take the derivative:
$f(x) = 2x^5 + 8x - 7 \Rightarrow f'(x) = 10x^4 + 8$
As even powers are always positive, the derivative is always positive as well. If $f'(x)$ is always positive, $f(x)$ can only cross the x-axis once and there's only one real root.
