Find the maximum of $x$ satisfies the cubic equation Let $a,b,c,d$ be integers with $1\leq a,b,c,d\leq10$, and $x$ be real such that
$ax^3-bx^2-cx-d=0$. How to find the maximum of $x$ ?
Thanks in advance.
 A: From Descartes' rule of signs, it is evident the cubic $f(x) = ax^3-bx^2-cx-d$ has exactly one positive root for all allowable $a, b, c, d$, so we just need to find its maximum value.  
Let $r$ be the positive root of $p(r)=r^3-10(r^2+r+1)$. It is clear $p(x)> 0$ for $x> r$, as $r$ is the only one positive root, where the sign turns from negative to positive.
Now if $x>r$, we have $f(x)\ge p(x)>p(r)=0$.  Thus $f(x)$ cannot have any root $> r$.  
Numerically $r \approx 10.993$, but if an exact formula is needed, you need to use a method like Cardano's formula to get $r = \frac13 \left(10+\sqrt[3]{5 (317+3 \sqrt{1401})}+\sqrt[3]{5(317-3 \sqrt{1401})}\right)$.
A: This is to find the largest zero of above polynomial. The estimate can be found, when coefficients are not known, by using  zero bounds for polynomials. For example if we use Cauchy's zero bound for the location of the zeros of a polynomial, we shall get $|x|\leq 1+\frac{1}{a}\max\left\{b,c,d\right\}\leq 11.$
A: discriminant of this equation is 
$c^2b^2-4(db^3+c^3a)+18abcd-27d^2b^2$ thus when the  discriminant will have max value that equation will have max value for $x$.or use vieta's formula
