1. **State the problem:** Factor the polynomial $3x^2 + 2x - 5$ fully. 2. **Recall the factoring method for trinomials when the leading coefficient is not 1:** We look for two numbers that multiply to $a \times c$ and add to $b$, where the polynomial is $ax^2 + bx + c$. 3. For $3x^2 + 2x - 5$, we have $a=3$, $b=2$, and $c=-5$. Calculate $a \times c = 3 \times (-5) = -15$. 4. Find two numbers that multiply to $-15$ and add to $2$. These numbers are $5$ and $-3$ because $5 \times (-3) = -15$ and $5 + (-3) = 2$. 5. Rewrite the middle term using these numbers: $$3x^2 + 5x - 3x - 5$$ 6. Group terms: $$(3x^2 + 5x) + (-3x - 5)$$ 7. Factor each group: $$x(3x + 5) - 1(3x + 5)$$ 8. Factor out the common binomial: $$(3x + 5)(x - 1)$$ **Final answer:** $$3x^2 + 2x - 5 = (3x + 5)(x - 1)$$