1. **State the problem:** Simplify or factor the quadratic expression $15x^2 - 6x - 5$. 2. **Recall the factoring formula:** For a quadratic $ax^2 + bx + c$, we look for two numbers that multiply to $a \times c$ and add to $b$. 3. **Calculate the product and sum:** Here, $a=15$, $b=-6$, and $c=-5$. So, $a \times c = 15 \times (-5) = -75$ and $b = -6$. 4. **Find two numbers that multiply to $-75$ and add to $-6$:** These numbers are $-15$ and $5$ because $-15 \times 5 = -75$ and $-15 + 5 = -10$ (not $-6$). So, try other pairs. Try $-10$ and $7.5$ (not integers), so let's try factoring by grouping directly. 5. **Factor by grouping:** Rewrite the middle term $-6x$ as $-15x + 9x$ (since $-15 + 9 = -6$). $$15x^2 - 15x + 9x - 5$$ 6. **Group terms:** $$(15x^2 - 15x) + (9x - 5)$$ 7. **Factor out common factors:** $$15x(x - 1) + 1(9x - 5)$$ Since the binomials are not the same, this approach fails. 8. **Use the quadratic formula to find roots:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 15 \times (-5)}}{2 \times 15}$$ $$= \frac{6 \pm \sqrt{36 + 300}}{30} = \frac{6 \pm \sqrt{336}}{30}$$ $$= \frac{6 \pm 4\sqrt{21}}{30} = \frac{3 \pm 2\sqrt{21}}{15}$$ 9. **Write the factorization using roots:** $$15x^2 - 6x - 5 = 15 \left(x - \frac{3 + 2\sqrt{21}}{15}\right) \left(x - \frac{3 - 2\sqrt{21}}{15}\right)$$ or equivalently, $$= (15x - (3 + 2\sqrt{21}))(15x - (3 - 2\sqrt{21}))$$ **Final answer:** The quadratic does not factor nicely over integers, but its roots are $x = \frac{3 \pm 2\sqrt{21}}{15}$.