1. **State the problem:** Solve $a^3+a^2=36$ for $a$. 2. **Write the equation in standard form:** $$a^3+a^2-36=0$$ 3. **Look for a simple integer solution by testing factors of $36$:** Since the constant term is $-36$, try $a=3$. $$3^3+3^2=27+9=36$$ So $a=3$ is a solution. 4. **Factor the polynomial using $a=3$:** $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ Check by expanding: $$ (a-3)(a^2+4a+12)=a^3+4a^2+12a-3a^2-12a-36=a^3+a^2-36 $$ 5. **Set each factor equal to zero:** $$a-3=0 \quad \text{or} \quad a^2+4a+12=0$$ So one solution is: $$a=3$$ 6. **Check the quadratic for real solutions:** $$a^2+4a+12=0$$ Use the discriminant: $$b^2-4ac=4^2-4(1)(12)=16-48=-32$$ Because the discriminant is negative, the quadratic has no real roots. 7. **Final answer:** $$a=3$$