<s>[INST] $x^3-9x-5=0$, then what is $x^4-18x^3-81x^2-12$ If we have $x^3-9x-5=0$, then what $x^4-18x^3-81x^2-12$ equals to?
This is a multiple choices question. 
A)$5$ B)$25$ C)$42$ D)$67$ E)$81$.
My attempt, 
By long division of polynomials, we have $x^4-18x^3-81x^2-12=(x-18)(x^3-9x-5)+(-72x^2-157x-102)$
Since $x^3-9x-5=0$, whatever $x$ could be, $x^4-18x^3-81x^2-12=-72x^2-157x-102$ 
How to proceed? 
P/S. Conclusion, NONE of the options can be right.
 [/INST] Look at the decomposition
$$x^4-18x^3-81x^2-12=x^2(x^2-18x-81)-12$$
Observe the roots of the equation  $x^3-9x-5=0$ must belong to the following two intervals: $(-3,-2)$ and $(3,4)$. 
The axis of symmetry of this function $f=x^2-18x-81$ is $x=9$, so the trend of the function is decreasing for $x < 9$, so observe:
$$\begin{array} {cccc} f(-3)<0&f(-2)<0&f(3)<0&f(4)<0\end{array}$$
So the function $f=x^2-18x-81<0$ in the interval of $(-3,-2)$ to $(3,4)$.
Since $x^2>0$ and $-12<0$, then $x^2(x^2-18x-81)-12$ must be less than zero. Since all the given options are positive, none of them can be right.
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