1. **State the problem:** Factor the trinomial $$2z^2 - 7z - 15$$ into the form $$(2z + A)(z + B)$$ where $A$ and $B$ are constants to be found. 2. **Recall the factoring method for trinomials with leading coefficient not 1:** - Multiply the leading coefficient and the constant term: $$2 \times (-15) = -30$$. - Find two numbers that multiply to $$-30$$ and add to the middle coefficient $$-7$$. 3. **Find the two numbers:** - The pair is $$3$$ and $$-10$$ because $$3 \times (-10) = -30$$ and $$3 + (-10) = -7$$. 4. **Rewrite the middle term using these numbers:** $$2z^2 + 3z - 10z - 15$$ 5. **Group terms and factor by grouping:** $$ (2z^2 + 3z) + (-10z - 15) $$ $$ z(2z + 3) - 5(2z + 3) $$ 6. **Factor out the common binomial:** $$ (z - 5)(2z + 3) $$ 7. **Rewrite in the requested form:** $$(2z + 3)(z - 5)$$ 8. **Answer:** The missing terms are $$3$$ and $$5$$, so the factorization is $$(2z + 3)(z - 5)$$. This means the blanks in $$(2z + [?])(z - [?])$$ are $$3$$ and $$5$$ respectively.