Find $t$ such that $x-3$ is a factor of $x^3-3x^2+tx+27$.
I know that if $x-3$ is a factor of a polynomial, then the polynomial must be zero when $x=3$. So I can plug in $x=3$ and set the polynomial equal to zero, and then solve for $t$. That gives me $3^3-3(3)^2+t(3)+27=0$. Simplifying, I get $27-27+3t+27=0$, or $3t+27=0$. Subtracting 27 from both sides, I get $3t=-27$. Dividing by 3, I get $t=-9$.

