Question

Find the zeros of the function y= (x-2)(x+1)(x+3)

Answer 1

$\bf y=(x-2)(x+1)(x+3)\implies 0=(x-2)(x+1)(x+3) \\\\\\ \begin{cases} 0=x-2\implies &~~2=x\\ 0=x+1\implies &-1=x\\ 0=x+3\implies &-3=x \end{cases}$