1. State the problem: Solve $a^3+a^2=36$ for $a$. 2. Set up the equation in standard form: $$a^3+a^2-36=0$$ 3. Factor by grouping: $$a^3+a^2-36=a^2(a+1)-36$$ 4. Find factors of $36$ that can fit $a^2(a+1)$ by trying integer roots (Rational Root Theorem). 5. Test $a=3$: $$3^3+3^2=27+9=36$$ 6. Conclude $a=3$ is a root, so $(a-3)$ is a factor of $a^3+a^2-36$. 7. Divide to factor the cubic: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ 8. Set each factor equal to zero: 1) $$a-3=0$$ 2) $$a^2+4a+12=0$$ 9. Solve $a-3=0$: $$a=3$$ 10. Solve the quadratic using the discriminant: $$a^2+4a+12=0 \Rightarrow \Delta=4^2-4\cdot 1\cdot 12=16-48=-32$$ 11. Since $\Delta<0$, the quadratic has no real solutions (it has complex solutions). 12. Final answer: - Real solution: $a=3$. - Complex solutions (optional): $$a=-2\pm 2i\sqrt{2}$$