# Prove $x^5+10x^3+ax^2+bx+c=0$ has no more than four real roots

I'm trying to prove that $$f(x) = x^5+10x^3+ax^2+bx+c=0$$ can not have more than four real roots, no matter the values of $a,b,c$ real Numbers .

My attempt: $f'(x) = 5x^4+30x^2+2ax+b =0$ and $f''(x) = 20x^3+60x+2a$, now $f(x)$ is differentiable and continuous in all real line but from here I want to use the intermediate value theorem, however I don't know how to apply it for $f$.

• Are $a, b, c$ real or complex? If they are real, then the quintic can have either 5, 3, or 1 real roots because any polynomial with real coefficients always has an even number of complex roots. Apr 18 '16 at 16:03
• Are real Numbers Apr 18 '16 at 16:12
• Actually you don't need to specify the coefficients $a,b,c$ as real. If any of them has a nonzero imaginary part, then $f(x)$ has at most two real roots. Apr 18 '16 at 16:18

$f'''(x)=60x^2+60$ has no real roots

$\implies f''$ has at most $1$ real root

$\implies f'$ has at most $2$ real roots

$\implies f$ has at most $3$ real roots

This argument uses Rolle's theorem at each step.

• Do you apply Rolle's theorem on f? Or f''' Apr 18 '16 at 16:13
• @Knight, if $f''$ had at least $2$ real roots, then $f'''$ would have at least $1$ real root. Repeat.
– lhf
Apr 18 '16 at 16:29