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Is there a (theoretical) method to tell the number of real roots of $2x^5+8x-7=0?$

Note, without using calculator.

3 Answers 3

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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.

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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 $1$.

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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?