jmc

We can factor the expression $x-3$ out of each term: $$2x(x-3)+3(x-3)=2x\cdot (x-3)+3\cdot (x-3)=\boxed{(2x+3)(x-3)}.$$ If you don't quite see how this works, suppose we put $A$ in place of $x-3$ everywhere in the original expression. Then we can see the factoring more clearly: $$2xA+3A=2x\cdot A+3\cdot A=(2x+3)A.$$ Putting $x-3$ back in for $A$, we have our factorization: $(2x+3)(x-3)$.