Question

Let $A=\{x:(x-1\left)\right(x^2-10+21)=0\}$ and $B=\{y:y\text{is a prime factor of}42\}$, then the number of common elements between $A\times B$ and $B\times A$ is

Answer 1

$A=\{x:(x-1\left)\right(x^2-10+21)=0\}$
$(x-1)(x^2-10+21)=0$
$\Rightarrow (x-1)(x^2-7x-3x+21)=0$
$\Rightarrow (x-1)(x-7)(x-3)=0$
$\Rightarrow x=1,3,7$

$B=\{y:y\text{is a prime factor of}42\}$
Prime factors of $42$ is $2,3,7$

Common elements between $A$ and $B$ is $2.$
$\therefore$ The number of common elements between $A\times B$ and $B\times A$ is $2^2=4$