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**running-gag** $\displaystyle f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

$\displaystyle f'(X)=4X^3 + 6X^2 + 8X + 2$

$\displaystyle f"(X)=12X^2 + 12X + 8 = 4(3X^2 + 3X + 2)$ This is the derivative isn't it ?

$\displaystyle f"(X) \geq 0$ therefore f' is increasing over R and f' has one real root $\displaystyle \alpha$ Why are you taking the derivative of a derivative ?

f' is negative over $\displaystyle ]-\infty,\alpha]$ and positive over $\displaystyle [\alpha,-\infty[$

f is decreasing over $\displaystyle ]-\infty,\alpha]$ and increasing over $\displaystyle [\alpha,-\infty[$

We just need to prove that $\displaystyle f(\alpha) > 0$

$\displaystyle f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

$\displaystyle f(X)=\left(\frac12 X + \frac14\right)f'(X) + \left(\frac54 X^2 + \frac12 X+\frac14\right)$

$\displaystyle f(\alpha)=\frac54 \alpha^2 + \frac12 \alpha + \frac14$ which is positive because the discriminant is negative