Question

Let $Z$ be the set of integers. If $A=\{x\in Z:e^{x^3-4x^2-7x+10}=1\}$ and $B=\{x\in Z:(1-x^2\left)\right(x^2-2x-8)\ge 0\},$ then $n\left[\right(A\cup B)\times (A\cap B\left)\right]$ is

Answer 1

$e^{x^3-4x^2-7x+10}=1=e^0$
$\Rightarrow x^3-4x^2-7x+10=0$
By trial and error method, we can find that $x=1$ is a solution.

Applying synthetic division
$\begin{array}{ccccc}x=1& 1& -4& -7& 10\\ & 0& 1& -3& -10\\ & 1& -3& -10& 0\end{array}$

$\Rightarrow x^3-4x^2-7x+10=(x-1)(x^2-3x-10)=0$
$\Rightarrow (x-1)(x-5)(x+2)=0$
$\Rightarrow x=-2,1,5$
$\Rightarrow A=\{-2,1,5\}$

$B=\{x\in Z:(1-x\left)\right(1+x\left)\right(x-4\left)\right(x+2)\ge 0\}$
$\Rightarrow B=\{-2,-1,1,2,3,4\}$

$A\cup B=\{-2,-1,1,2,3,4,5\}$
$A\cap B=\{-2,1\}$

$n\left[\right(A\cup B)\times (A\cap B\left)\right]$
$=n(A\cup B)\times n(A\cap B)=7\times 2=14$