Number of real solutions to $x^7 + 2x^5 + 3x^3 + 4x = 2018$

Question

Find the number of real solutions of $x^7 + 2x^5 + 3x^3 + 4x = 2018$?

What is the general approach to solving this kind of questions? I am interested in the thought process.

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Few of my thoughts after seeing this question: since $x$ has all odd powers so, it can not have any negative solution. 2018 is semiprime; not much progress here. We can sketch the curve but graphing a seven order polynomial is difficult.

Answer 1

If $x\le 0$ the left hand side is negative therefore no solution. We suppose $x>0$ and we consider $f(x)=x^7+2x^5+3x^3+4x$, then $f$ is the sum of increasing functions therefore increasing. Since $f(0,\infty)=(0,\infty)$ this equation has only one solution.

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This can be done by differentiation which give a more simple proof since the derivative is clearly positive.

Answer 2

Render

$x^7+0x^6+2x^5+0x^4+3x^3+0x^2+4x-2018=0$

Descartes' Rule of Signs forces exactly one positive root and no negative roots. That's it!