Both $(x-1)$ and $(x+1)$ are factors of $p(x)=x^3+a{x}^2+bx+6$. Find a and $b$.

**$(x-a)$ is a factor of $p(x)$ if and only if $p(a)=0$.**

It's a special case of the remainder theorem (with remainder 0).

Use it to test factors:
Is $(x+3)$ a factor of $p(x)=x^2+5x+6$?
Substitute $x=-3$ (note: $x+3=0$ → $x=-3).p(-3)=9-15+6=0$. ✓ Yes, $(x+3)$ is a factor.

This is the standard way to factor cubics — guess and test small zeros.