### Instruction: Find $t$ such that $x-3$ is a factor of $x^3-3x^2+tx+27$.### Response: 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 $3t=-27$.