Two roots of the polynomial $x^4+x^3-7x^2-x+6$ are $2$ and $-3$. Find two other roots. I have divided this polynomial first with $(x-2)$ and then divided with $(x+3)$ the quotient. The other quotient I have set equal to $0$ and have found the other two roots. Can you explain to me if these actions are correct and why?

$(x-2)(x+3)=x^2+x-6$
$x^4+x^3-7x^2-x+6\\=(x^4+x^3-6x^2)-(x^2+x-6)\\= x^2(x^2+x-6)-1(x^2+x-6)\\=(x^2-1)(x^2+x-6)$
