1. **State the problem:** Factor fully the expression $3xy + 3xz + 2y + 2z$. 2. **Group terms:** Group terms to find common factors: $$3xy + 3xz + 2y + 2z = (3xy + 3xz) + (2y + 2z)$$ 3. **Factor out common factors in each group:** $$= 3x(y + z) + 2(y + z)$$ 4. **Factor out the common binomial factor:** $$= \cancel{(y + z)}(3x + 2) + \cancel{(y + z)}$$ $$= (y + z)(3x + 2)$$ 5. **Final answer:** The fully factored form is $$\boxed{(y + z)(3x + 2)}$$ This uses the distributive property and factoring by grouping.